Problem
Suppose there is a surface $F(x,y,z)=0$. Also $z=f(x,y)$. Substituting $z$ we get $F(x,y,f(x,y))=0$ for all $(x,y)$ in some open set $S$. Now we introduce an auxiliary function $g$ defined on $S$ as follows: $$ g(x,y)= F(x,y,f(x,y)). $$ So $g(x,y)=0$ on $S$. Hence partial derivatives of $g$ are also zero.(From Apostal Calculus volume 2)
I am not able to understand the last line. If the partial derivatives are zero, the gradient is always zero, which is not the case. For example let $F(u,v)$, where $u=xy$ and $v=(x^2+z^2)^{1/2}$. Partial derivatives exist at $(1,1,3)$.