The question is: 'Show that if f(x,y) is harmonic, then $f(x^2-y^2,2xy)$ is also harmonic using Laplace's equation: $\frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} = 0 $.
I end up with this using the chain rule: $$\frac{\partial^2f}{\partial x^2} = 2\frac{\partial f}{\partial u} + 4x^2\frac{\partial^2f}{\partial u^2} + 8xy\frac{\partial^2f}{\partial u \partial v} + 4y^2\frac{\partial^2f}{\partial v^2} $$
and
$$\frac{\partial^2f}{\partial y^2} = -2\frac{\partial f}{\partial u} + 4y^2\frac{\partial ^2f}{\partial u^2} + 4x^2\frac{\partial ^2f}{\partial v^2} - 8xy\frac{\partial^2f}{\partial u \partial v}$$
so: $$\frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} = (4x^2+4y^2)\frac{\partial^2f}{\partial u^2} + (4x^2+4y^2)\frac{\partial^2f}{\partial v^2}$$
which is not equal to $0$ as far as I can tell. I thought you might be able to say $Fuu + Fvv = 0$ just because it's harmonic, but I think you are meant to be showing that is harmonic.
Any ideas or advice is much appreciated.
It is equal to zero as
$$ \begin{split} \frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} &= (4x^2+4y^2)\frac{\partial^2f}{\partial u^2} + (4x^2+4y^2)\frac{\partial^2f}{\partial v^2}\\ &= (4x^2+4y^2)\left(\frac{\partial^2f}{\partial u^2} + \frac{\partial^2f}{\partial v^2} \right) \\ &=0 \end{split}$$
as $f(u, v)$ is harmonic.