Show that $s\frac{\partial u}{\partial s}-t\frac{\partial u}{\partial t}=2(s^2+t^2)\frac{\partial f}{\partial x}$ where $u=f(x,y) ; x=s^2-t^2,y=2st$

Question

I want to obtain the result

$$s\frac{\partial u}{\partial s}-t\frac{\partial u}{\partial t}=2(s^2+t^2)\frac{\partial f}{\partial x}$$

where, $$u=f(x,y) \text{ and } x=s^2-t^2,y=2st$$

So far I obatined, $$s\frac{\partial u}{\partial s}-t\frac{\partial u}{\partial t}=2(s^2+t^2)$$

To get this I used, $$\frac{\partial u}{\partial s}=\frac{\partial x}{\partial s}+\frac{\partial y}{\partial s}$$ and $$\frac{\partial u}{\partial t}=\frac{\partial x}{\partial t}+\frac{\partial y}{\partial t}$$

I don't know how to obtained the desired results.
Am I doing something wrong?
If so what is te correct way of doing this?

Answer 1

The correct derivatives should be, $$\frac{\partial u}{\partial s}=\frac{\partial x}{\partial s}\frac{\partial f}{\partial x}+\frac{\partial y}{\partial s}\frac{\partial f}{\partial y}$$

and, $$\frac{\partial u}{\partial t}=\frac{\partial x}{\partial t}\frac{\partial f}{\partial x}+\frac{\partial y}{\partial t}\frac{\partial f}{\partial y}$$