Usage of implicit function theorem for $f(x,y)=x^2+2xy-y^2-a^2$

Question

Find the derivative of the following implicit function with the implicit function theorem:

$$F(x,y)=x^2+2xy-y^2-a^2$$

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My attempt for this task: $$F(x,y)=0 \Leftrightarrow (x,y)=(a,0)$$

Derivate partial: $$\partial_x F(x,y)=2x+2y\\ \partial_y F(x,y)=2x-2y$$

Prove, if $\partial_y$ invertible: $$\partial_y \text{invertible} \, \Leftrightarrow a \neq 0$$

So I get for the derivate: $$f'(x_0,y_0)=-\left(\frac{x+y}{x-y}\right)$$

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Is this the right and easiest way?

Answer 1

We have $$F_x(x,y)=2(x+y),\qquad F_y(x,y)=2(x-y).$$ Thus as long as $F_y(x,y)\neq 0$, i.e. $x\neq y$ we have by the implicit function theorem: $$\frac{d y}{dx}=-\frac{F_x(x,y)}{F_y(x,y)}=-\frac{2(x+y)}{2(x-y)}=\frac{x+y}{y-x}.$$

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Note in cases where $x=y$, we can instead find the derivative of $x$ with respect to $y$. As long as $F_x(x,y)\neq 0$, i.e. $x\neq -y$

$$\frac{d x}{dy}=-\frac{F_y(x,y)}{F_x(x,y)}=-\frac{2(x-y)}{2(x+y)}=\frac{y-x}{x+y}.$$

Answer 2

i have got $$y'=\frac{x+y}{y-x}$$ and we get the condition $$x\ne y$$