partial derivative with changing variables Here is the question

and the way that I attempted for the first part in the question

All of them seems like correct way but I don't understand what is the relationship of the functions $$F(u,v)\quad and\quad f(x,y)$$ 
and why it can establish
\begin{align}
\frac{{\partial F}}{{\partial u}}=\frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}}+\frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}
\end{align}
and
\begin{align}
\frac{{\partial F}}{{\partial v}}=\frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial v}}+\frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial v}}
\end{align}
In the second part, I don't understand what does $$ f(x,y) \text{ is independent of } x$$
mean?
Please help!!
 A: $x=u^2-v^2, y=2uv$. So $$\frac{{\partial F}}{{\partial u}}=\frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}}+\frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}
=2u\frac{{\partial f}}{{\partial x}}+2v\frac{{\partial f}}{{\partial y}}$$
Similarly \begin{align}
\frac{{\partial F}}{{\partial v}}=\frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial v}}+\frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial v}}=-2v\frac{{\partial f}}{{\partial x}}+2u\frac{{\partial f}}{{\partial y}}
\end{align}
Square and subtract. What do you get ?
\begin{align}\left(\frac{{\partial F}}{{\partial u}}\right)^2+\left(\frac{{\partial F}}{{\partial v}}\right)^2&=\left(2u\frac{{\partial f}}{{\partial x}}+2v\frac{{\partial f}}{{\partial y}}\right)^2+\left(-2v\frac{{\partial f}}{{\partial x}}+2u\frac{{\partial f}}{{\partial y}}\right)^2\\
&= 4(u^2+v^2)\left[\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2\right]
\end{align}
as you have done. As to the next part,
\begin{align}u\frac{{\partial F}}{{\partial u}}-v\frac{{\partial F}}{{\partial v}}&=2u^2\frac{{\partial f}}{{\partial x}}+2uv\frac{{\partial f}}{{\partial y}}-2v^2\frac{{\partial f}}{{\partial x}}-2uv\frac{{\partial f}}{{\partial y}}\\&=2(u^2-v^2)\dfrac{\partial f}{\partial x}\end{align}
So $$u\frac{{\partial F}}{{\partial u}}-v\frac{{\partial F}}{{\partial v}}=0\implies 2(u^2-v^2)\dfrac{\partial f}{\partial x}=0\implies2x\dfrac{\partial f}{\partial x}=0$$
Since $x\ne 0$. This means $$\frac{\partial f}{\partial x}=0\implies f(x,y)=g(y)$$
which shows that $f(x,y)$ is independent of $x$.
