Hello Chokfull

I think you also missed out a power of 2. The equation should be$\displaystyle (x^2+y^2)^2 = 25(x^2-y^2)$

So, when we differentiate:$\displaystyle 2(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)=50x -50y\frac{dy}{dx}$

and putting $\displaystyle \frac{dy}{dx}=0$ gives:$\displaystyle x^2+y^2 = \frac{25}{2}$ (I think - not $\displaystyle \frac{25}{4}$)

Now we put this back into the original equation, using the hint that $\displaystyle x^2-y^2 = (x^2+y^2)-2y^2$:$\displaystyle \left(\frac{25}{2}\right)^2=25\left(\frac{25}{2}-2y^2\right)$

which you can now solve for $\displaystyle y$, and hence for $\displaystyle (x,y)$.

Grandad