One more time, z IS NOT the position of the bug, z is the distance of the observer from the plane containing the bug.

And? $$x$$ and $$y$$ are also not the bug's coordinates. They're general coordinates on the plane.

For the purpose of calculating the bug's temperature from $$T(t,\,x,\,y,\,z)$$, you have to set $$x(t) = f(t)$$, $$y(t) = g(t)$$, and $$z(t) = 0$$. Then, and

**only** then, are you taking a total derivative.

There is a difference between saying

$$

\frac{\mathrm{d}T}{\mathrm{d}t} \,=\, \frac{\partial T}{\partial t} \,+\, \frac{\partial T}{\partial x} \, \frac{\mathrm{d}x}{\mathrm{d}t} \,+\, \frac{\partial T}{\partial y} \, \frac{\mathrm{d}y}{\mathrm{d}t}

$$

because $$\frac{\mathrm{d}z}{\mathrm{d}t} = 0$$, which is a total derivative, and

$$

\Bigl( \frac{\partial T}{\partial t} \Bigr)_{z} \,=\, \frac{\partial T}{\partial t} \,+\, \frac{\partial T}{\partial x} \, \frac{\mathrm{d}x}{\mathrm{d}t} \,+\, \frac{\partial T}{\partial y} \, \frac{\mathrm{d}y}{\mathrm{d}t}

$$

(for lack of better notation), because you're leaving $$z$$ independent, in which case you're still taking a partial derivative, specifically of the function

$$

(t,\,z) \,\mapsto\, T(t,\, x(t),\, y(t),\, z) \,.

$$

To calculate the bug's temperature from $$T$$, you have to set $$z(t) = 0$$, and you're therefore in the former case. And you're calculating a total derivative

**only** because you set $$z(t) = 0$$.

Do you understand this? Because it's the point we've been trying to get through to you.