At least, prometheus and rpenner stopped arguing, once he realized that they he might be on the wrong side of the argument.
You have no basis on which to assert that prometheus and I have
stopped arguing and certainly no basis to assert what realizations we might have have.
I believe, for reasons of finite resources requiring us to work or sleep or spend times with friends and family, that both of us independently have come to the conclusion that your immediate education is not our number one priority at all hours of the day and night.
I see that you have removed my name from this sentence on the basis of data.
--
So what have we learned?
If $$f(x_1, ..., x_n)$$ is a function then the total derivative with respect to one of it's inputs
$$\frac{d \quad}{d x_j} f = \sum_{i=1}^n \frac{\partial f}{\partial x_i} \times \frac{d x_i}{d x_j} = \frac{\partial f}{\partial x_j} + \sum_{i \neq j } \frac{\partial f}{\partial x_i} \times \frac{d x_i}{d x_j} $$
And that this looks like the inner product of $$\left( \vec{\nabla} f \right) \cdot \begin{pmatrix} \frac{d x_i}{d x_1} \\ \vdots \\ \frac{d x_i}{d x_{j-1}} \\ \rule[-1em]{0pt}{25em} 1 \\ \rule[-1em]{0pt}{25em} \frac{d x_i}{d x_{j+1}} \\ \vdots \\ \frac{d x_i}{d x_n} \end{pmatrix} =\begin{pmatrix} \rule[-10em]{0pt}{25em} \frac{\partial f}{\partial x_1} \\ \vdots \\ \rule[-10em]{0pt}{25em} \frac{\partial f}{\partial x_{j-1}} \\ \rule[-10em]{0pt}{25em} \frac{\partial f}{\partial x_j} \\ \rule[-10em]{0pt}{25em} \frac{\partial f}{\partial x_{j+1}} \\ \vdots \\ \rule[-10em]{0pt}{25em} \frac{\partial f}{\partial x_n} \end{pmatrix} \cdot \begin{pmatrix} \rule[-10em]{0pt}{25em} \frac{d x_i}{d x_1} \\ \vdots \\ \rule[-10em]{0pt}{25em} \frac{d x_i}{d x_{j-1}} \\ \rule[-10em]{0pt}{25em} 1 \\ \rule[-10em]{0pt}{25em} \frac{d x_i}{d x_{j+1}} \\ \vdots \\ \rule[-10em]{0pt}{25em} \frac{d x_i}{d x_n} \end{pmatrix} $$
But where post 18 jumps the shark is where it goes from talking about an abstract function of variables, to a concrete function where the only variables seen are $$\theta$$ and $$x$$. There is no formal dependency on u and v in that three-part sum.
In $$f(x,y,z) = xyz$$, $$\frac{d f}{d x} = \frac{d \quad}{d x} xyz = yz + xz \frac{dy}{dx} + xy \frac{dz}{dy}$$ but if you change the function under discussion to $$f = 3xt$$ I don't have any logical basis to think you are talking about $$f(x,t,3)$$ or $$f(x,3t,1)$$. Nevertheless, the partial derivative of 3xt with respect to x is the same no matter how this expression relates to the original f. $$\frac{\partial \quad}{\partial x} 3xt = 3t$$. Math is both rigid and fragile like that so the speaker has to be precise.
The term partial in "partial derivative" does not mean you have carte blanche to stop the process of calculating the derivative at any intermediate step. It means that the derivative is only with repect to changing one variable, and holding the others constant. When you confuse matters by first saying $$f(\theta,u,v) = 3 \theta + u + v$$ and then talking about the case which was constrained to $$u = \sin^2 \theta, \; v = \ln x$$ (this is hypothetical, because post 18 isn't this explicit) then you are no longer talking about the same f. This new function, for which I shall use a new name, is $$g(\theta, x) = 3 \theta + \sin^2 \theta + \ln x$$.
As should be clear, the partial derivatives of f and g look nothing alike.
$$\frac{\partial f}{\partial \theta} = 3 , \; \frac{\partial f}{\partial u} = 1 , \; \frac{\partial f}{\partial v} = 1 , \; \frac{\partial f}{\partial x} = 0 \\ \frac{\partial g}{\partial \theta} = 3 + 2 \sin \theta \, \cos \theta , \; \frac{\partial g}{\partial u} = 0 , \; \frac{\partial g}{\partial v} = 0 , \; \frac{\partial g}{\partial x} = x^{\tiny -1} \\ \frac{\partial u}{\partial \theta} = 2 \sin \theta \, \cos \theta , \; \frac{\partial u}{\partial u} = 1 , \; \frac{\partial u}{\partial v} = 0 , \; \frac{\partial u}{\partial x} = 0 \\ \frac{\partial v}{\partial \theta} = 0 , \; \frac{\partial v}{\partial u} = 0 , \; \frac{\partial v}{\partial v} = 1 , \; \frac{\partial v}{\partial x} = \frac{1}{x} $$
Post 18 is fundamentally flawed when it skips steps of defining u and v and when it skipped the step of defining f the way you meant it defined. Because of these miscommunications, post 18 is indefensibly wrong, because you flipflop between a definition of f that is written for us and a definition that exists only in your head in some nebulous form and would not begin to be expressed for us until post 27. But instead of fixing the problem, you dug yourself in deeper with misstating the definition of partial differentiation and perversely ignoring the obvious theta-dependency of the second term.
Partial differentiation requires a formal expression defined over a certain set of variables. When you write a nebulous expression like $$f = 3 \theta + \sin^2 \theta + \ln x$$ then your readers assume that $$theta$$ and $$x$$, being otherwise unexplained, are among the variables. Strictly formal rules (the sum rule, the product rule, the law of composition, the chain rule, ...) allow us to process the expression. Alternately, you can go the route AlphaNumeric kindly went and explicitly compute the partial derivative.
But just as f and g had different definitions in your head (but were both written as f in post 18) so do f and g have different definitions of parameter space. This is not surprising because if $$f(\theta,u,v)$$ has a 3-dimensional parameter space, the constraint $$u = \sin^2 \theta$$ reduces it to a 2-dimensional parameter space of independent variables $$\theta$$ and v. You have eliminated all the posibilities to consider when $$u /neq \sin^2 \theta$$ and this is part of the reason why $$f(\theta,u,v)$$ cannot be considered identical to $$g(\theta, x)$$ and if f and g are not identical, there is no reason to think that there partial derivatives must be.