Given $z^2=x^2+y^2$ find $z_x$ at $(x,y,z)=(3,4,5).$
This is how I went about the problem, but I don't think it's right.
$z^2=x^2+y^2 \Longrightarrow f(x,y,z)=x^2+y^2-z^2$ so $f_x(x,y,z)=2x$ and $f_x(3,4,5)=6.$
Again, I don't think this is right because I didn't make use of the $4$ or $5$ anywhere.
Since we are dealing with $P=(3,4,5)$, we have that
$$z^2=x^2+y^2 \implies z(x,y)=\sqrt{x^2+y^2}$$
and then
$$z_x(x,y)=\frac{x}{\sqrt{x^2+y^2}}\implies z_x(3,4)=\frac35$$