When I take the first partial derivative of this expression with respect to $x$, I get this (with help of the quotient rule):
The function is as follows: $$f(x,y)=\dfrac{x-y}{x+y}$$
1st partial derivative with respect to $x$: $$\dfrac{2y}{x^2+2xy+y^2}$$
Why can't I use the quotient rule again, if I want to find the 2nd partial derivative with respect to $x$? The solution told me I have to take $2y$ out of the expression and treat it as a constant and then solve for $1/(x^2+2xy+y^2)$.
Let $$f(x,y)=\frac{x-y}{x+y}$$ Then, if we derive w.r.t. x, we get $$\frac{\partial f}{\partial x}(x,y)=\frac{2y}{(x+y)^2}$$ I think both methods are equivalent:
Quotient rule: knowing that the derivative of $2y$ w.r.t. $x$ is $0$ $$\frac{\partial^2f}{\partial x^2}(x,y)=\frac{\partial_{x}(2y)\cdot(x+y)^2-2y\cdot\partial_{x}((x+y)^2)}{(x+y)^4}=\frac{-2y\cdot 2(x+y)}{(x+y)^4}=\frac{-4y}{(x+y)^3}$$ where $\partial_{x}=\frac{\partial}{\partial x}$
The method you book suggests: since $2y$ does not depend on $x$, we can take out of the derivative w.r.t. x $$\frac{\partial^2f}{\partial x^2}(x,y)=\frac{\partial}{\partial x}\frac{\partial f}{\partial x}(x,y)=2y\frac{\partial}{\partial x}\frac{1}{(x+y)^2}=2y\frac{-1\cdot2(x+y)}{(x+y)^4} = \frac{-4y}{(x+y)^3}$$
So you can go both ways, thank to the fact that $2y$ does not depend on $x$.