If $f(x,y)$ is a harmonic function show that the function $F(x,y)=f(x^2-y^2,2xy)$ is also harmonic.
You have to use Laplace's formula to prove this, unless there is an easier way.
I'm having trouble figuring out how to properly write this implicitly in order to take the second partial derivative for each term.
On
If $f$ is harmonic on $\mathbb{R}^{2}$, then $f(x,y)=\Re F(x+iy)$ is the real part of a function $F$ which is a holomorphic for $z=x+iy \in \mathbb{C}$. Notice that $x^{2}-y^{2}=\Re (z^{2})$ and $2xy=\Im (z^{2})$. So $f(x^{2}-y^{2},2xy) = \Re F(z^{2})$ is the real part of a holomorphic function and, therefore, is harmonic.
You can just do it out using the chain rule. Maybe some notation will help. Let $f^{(n,m)}(...,...)$ be the the $n$-th partial derivative with respect to the first argument and $m$-th partial with respect to the second argument. I've purposefully left the arguments unnamed since they don't have to actually be $x$ or $y$. If $f(x,y)$ is harmonic in $x$ and $y$ then \begin{equation} f^{(2,0)}(x,y) + f^{(0,2)}(x,y)=0~. \end{equation} The arguments don't matter here, so it's also true that \begin{equation} f^{(2,0)}(x^2-y^2,2xy) + f^{(0,2)}(x^2-y^2,2xy)=0~. \end{equation} Now you want to show that \begin{equation} \left({\partial^2\over\partial x^2}+{\partial^2\over\partial y^2}\right)F(x,y)=0 \end{equation} given that $f$ is harmonic. Using the chain rule, \begin{equation} {\partial \over \partial x}f(x^2-y^2, 2xy) = 2x f^{(1,0)}(x^2-y^2,2xy) + 2y f^{(0,1)}(x^2-y^2,2xy)~. \end{equation} I'll take it one step further and compute the second partial of the first term: \begin{eqnarray} {\partial \over \partial x} 2xf^{(1,0)}(x^2-y^2,2xy) &=& 2 f^{(1,0)}(x^2-y^2,2xy) \\&&+ 2x\left(2x f^{(2,0)}(x^2-y^2,2xy) + 2y f^{(1,1)}(x^2-y^2,2xy)\right) \end{eqnarray} You can apply the chain rule some more to fully evaluate the Laplacian yourself. In the end you'll get something proportional to \begin{equation} f^{(2,0)}(x^2-y^2,2xy) + f^{(0,2)}(x^2-y^2,2xy) \end{equation} which I've already argued vanishes.