Non zero partial derivatives in implicit function theorem?

Question

Simple proofs of this in 3 dimensions $z(x,y)$ impose a $dz = 0$ constraint to solve for $dy/dx$. This implies $z(x, y) = \text{constant}$. Why are the partial derivatives in general in the theorem non zero if $z = \text{constant}$?

Answer 1

The point is that the equation $z(x,y)=c$ holds only for some $y$ (dependent on $x$). For example, the equation $x^2+y^2=1$ holds for some $(x,y)$ and not others. The partial derivatives of the function $z(x,y)=x^2+y^2$ are not zero.

Then we consider $y(x)$, a function defined implicitly by the equation $z(x,y)=c$. It's true that $z(x,y(x))=c$ for all $x$. So, when we differentiate this composition with respect to $x$, the result is zero.

In the above example, $\frac{d}{dx}(x^2+y(x)^2 )=0$ if $y(x)=\sqrt{1-x^2}$. This does not contradict the fact that $x^2+y^2$ has nonzero partials.