So I want to solve the following question:
"Let $z=f(x,y)$ be a function of type $C^2$ in the plane and introduce $u=x^2-y^2$ och $v=2xy$. Transform the expression $$ \frac{1}{x^2+y^2}\left(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}\right) $$ to the new variables $u,v$. The answer is not allowed to include $x$ or $y$. "
To start with, I let: $$ \left\{ \begin{array}{c} u=u(x,y)=x^2-y^2 \\ v=v(x,y)=2xy\\ \end{array} \right. \; \;(1)$$ We have $$\frac{\partial u}{\partial x}=2x, \frac{\partial u}{\partial y}=-2y, \frac{\partial v}{\partial x}=2y, \frac{\partial v}{\partial y}=2x $$
Given a $C^2$-function $f(x,y)$ I let $g(u,v)$ be the function which you get from $f(x,y)$ by switching the variables from $(x,y)$ to $(u,v)$. This would mean that you get $f(x,y)$ from $g(u,v)$ by switching the variables back from $(u,v)$ to $(x,y)$, as such:
$$f(x,y)=g(u(x,y),v(x,y))=g(x^2-y^2,2xy)\; \; \; \;(2)$$
Now I calculate the partial derivatives of $f(x,y)$ from the relationship (2) with the help of the chain rule. I use partially derive both terms of (2), first with respect to $x$ and then with respect to $y$ and get
$$f'_x=g'_u \cdot \frac{\partial u}{\partial x}+g'_v \cdot \frac{\partial v}{\partial x}=g'_u \cdot 2x + g'_v \cdot 2y \; , \; \; \;(3)$$
$$f'_y=g'_u \cdot \frac{\partial u}{\partial y}+g'_v \cdot \frac{\partial v}{\partial y}=g'_u \cdot -2y + g'_v \cdot 2x \; \; \; \;(4)$$
After that I derive both terms of (3) with respect to $x$ and both terms of (4) with respect to $y$ which gives us
$$f''_{xx}=\frac{\partial}{\partial x}(f'_x)= \\=g'_u \cdot 2x + g'_v \cdot 2y + \left(g''_{uu}\frac{\partial u}{\partial x}+ g''_{uv}\frac{\partial v}{\partial x}\right) \cdot 2x+\left(g''_{vu}\frac{\partial u}{\partial x}+ g''_{vv}\frac{\partial v}{\partial x}\right) \cdot 2y=\\=g'_u \cdot 2x + g'_v \cdot 2y +\left(g''_{uu}\cdot 2x+g''_{uv}\cdot2y\right)\cdot 2x+\left(g''_{vu}\cdot 2x+g''_{vv}\cdot2y\right)\cdot 2y=\\=g'_u \cdot 2x + g'_v \cdot 2y+2g''_{uu}\cdot4x^2+g''_{uv}\cdot 8xy+g''_{vv}\cdot4y^2 \; \; \; \; \; \; \; \; \; \;(5)$$
respectively
$$f''_{yy}=\frac{\partial}{\partial y}(f'_y)= \\-2y\cdot\left(g''_{uu}\frac{\partial u}{\partial y}+ g''_{uv}\frac{\partial v}{\partial y}\right) + 2x\cdot\left(g''_{vu}\frac{\partial u}{\partial y}+ g''_{vv}\frac{\partial v}{\partial y}\right)=\\=-2y \cdot \left(g''_{uu}\cdot -2y+g''_{uv}\cdot2x\right)+ 2x \cdot\left(g''_{vu}\cdot -2y+g''_{vv}\cdot2x\right)=\\= 4y^2 \cdot g''_{uu}-2g''_{uv}\cdot 4xy+ g''_{vv}\cdot4x^2 \; \; \; \; \; \; \; \; \; \;(6)$$
But what am I supposed to do now? I am thinking that I should insert $(5), (6)$ and $(4)$ in the original equation. But I am not exactly sure how to do it. And am I correct in my calculations or do I need to to something else due to it being of type $C^2$?