I think I may be missing something here,
$f(x,y)=\left\{ \frac {xy(x^{2}-y^{2})}{x^{2}+y^{2}}\right.\quad(x,y)\neq (0,0)$
Let $X(s,t)= s\cos(\alpha)+t\sin(\alpha)$ and $Y(s,t)=-s\sin(\alpha)+t\cos(\alpha)$, where $\alpha$ is a constant, and Let $F(s,t)=f(X(s,t), Y(s,t))$. Show that
$ \left.\frac{\partial^2 F}{\partial s^2}\frac{\partial^2 F}{\partial t^2} - \left( \frac{\partial^2 F}{\partial s\partial t}\right)^2 = \left(\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2\right)\right| _{x=X(s,t),y=Y(s,t)} $
I decided to try and subsitute My $X(s,t)$ and $Y(s,t)$ into $f(x,y)$, however im just wondering if thre is an alternative approach as it gives a lot of terms, many thanks in advance.
I have gone away and had a think about the answer and still not sure where to put my best foot forward with it so:
$ \frac{\partial^2 F}{\partial s^2}=cos^{2}\alpha\frac{\partial^2 F}{\partial ^2X}$ $\Rightarrow$ $ \frac{\partial^2 F}{\partial t^2}=sin^{2}\alpha\frac{\partial^2 F}{\partial ^2X}$
Now using the fact that $\frac{\partial^2 F}{\partial ^2X}$ is equal to $\frac{\partial^2 f}{\partial ^2X} | _{x=X(s,t)}$ to calculate our $\frac{\partial^2 F}{\partial ^2X}$.
Now $\frac{\partial^2 f}{\partial ^2X}$= $ \frac{-4x^{4}y-20x^{2} y^{3}+8x 3y^{3}-4x y^{5}+4x^{5} y+10x^{3} y^{3}+6x y^{5}}{(x^{2}+y^{2})^{3}}$ hence do I make the subsitution here, seems to be far to many terms and havent even got to the RHS, many thanks in advance.