Calculating the partial derivative of the following 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.
 A: I might be wrong but actually the only helpful thing i see is that $X^2 + Y^2=s^2+t^2$,so in your $f(X,Y)$ you'll get:
$$
\begin{align}
f(X,Y)& =\frac{XY(X^2-Y^2)}{X^2+Y^2} \\  \\  \\
& =\frac{((t^2-s^2)\sin(a)\cos(a)+st(\cos^2(a)-\sin^2(a))(t^2-s^2)(\sin^2(a)-\cos^2(a))}{t^2+s^2}
\end{align}
$$
Alternatively you could use the chain rule where $\frac{\partial F}{\partial s}=\frac{\partial F}{\partial X}\frac{\partial X}{\partial s}=\cos(a)\frac{\partial F}{\partial X}$ where  $\frac{\partial X}{\partial s}=\cos(a)=ct$. The partial derivative $\frac{\partial F}{\partial X}$ is actually the same as $(\frac{\partial f}{\partial x})_{x=X}$, something you will compute anyway. So you will have $\frac{\partial^2 F}{\partial s^2}=\cos(a)\frac{\partial\frac{\partial F}{\partial X}}{\partial s}=\cos(a)\frac{\partial\frac{\partial F}{\partial X}}{\partial X}\frac{\partial X}{\partial s}=\cos^2(a)\frac{\partial\frac{\partial F}{\partial X}}{\partial X}=\cos^2(a)\frac{\partial^2F}{\partial X^2}$.
Likewise you can compute the other partial derivatives for F.
A: Let $f:\ (x,y)\mapsto f(x,y)$ be an arbitrary function and put $$g(u,v):=f(u\cos\alpha + v\sin\alpha, -u\sin\alpha+v \cos\alpha)\ .$$
Using the abbreviations
$$c:=\cos\alpha, \quad s:=\sin\alpha,\quad \partial_x:={\partial\over\partial x}, \quad\ldots$$
we have (note that $c$ and $s$ are constants) 
$$\partial_u=c\partial_x-s\partial_y, \quad \partial_v =s\partial_x+ c\partial_y\ .$$
It follows that
$$\eqalign{g_{uu}&\cdot g_{vv}-\bigl(g_{uv}\bigr)^2 \cr
&=(c\partial_x-s\partial_y)^2 f\cdot (s\partial_x+ c\partial_y)^2 f -\bigl((c\partial_x-s\partial_y)(s\partial_x+ c\partial_y)f\bigr)^2 \cr
&=(c^2 f_{xx}-2cs f_{xy}+s^2 f_{yy})(s^2 f_{xx}+2cs f_{xy}+c^2 f_{yy}) -\bigl(cs f_{xx}+(c^2-s^2) f_{xy}-cs f_{yy}\bigr)^2 \cr
&=\ldots =f_{xx}\, f_{yy} -\bigl(f_{xy})^2\ . \cr}$$
This shows that the stated identity is true for any function $f$ and not only for the $f$ considered in the original question.
There should be a way to prove this identity "from a higher standpoint", i.e., without going through this tedious calculation.
