I need to show this to use it in another problem:
Lets say you have $u: \mathbb{R}^2 \rightarrow \mathbb{R}=u(x,y)$.
Then you make a new function $z(x,y)=u(x,-y)$
I need to show that
$\frac{\partial z}{\partial y}(a,b)=-\frac{\partial u}{\partial y}(a,-b)$. Is this true? How can I show it? I mean that I evaluate it in the points $(a,b)$ and $(a,-b)$.
Use chain rule, let $f_1(x,y))=x$ and $f_2(x,y)=-y$ then $z(x,y)=u(f_1(x,y),f_2(x,y))$,so:
$$\frac{\partial z}{\partial y}(x,y)=\frac{\partial u}{\partial f_1}(f_1(x,y),f_2(x,y))\cdot\frac{\partial f_1}{\partial y}(x,y)+\frac{\partial u}{\partial f_2}(f_1(x,y),f_2(x,y))\cdot\frac{\partial f_2}{\partial y}(x,y)=\\=\frac{\partial u}{\partial f_1}(f_1(x,y),f_2(x,y))\cdot 0+\frac{\partial u}{\partial f_2}(f_1(x,y),f_2(x,y))\cdot(-1)=-\frac{\partial u}{\partial f_2}(f_1(x,y),f_2(x,y))=\\=-\frac{\partial u}{\partial f_2}(x,-y)=-\frac{\partial u}{\partial y} (x,-y)$$