I am asked to find ${y}'$ of $(x^{2}+y^{2})^{2} = a^{2}(x^{2}-y^{2})$ with $y(x)$ and $a$ as a positive constant, which is given in the solutions as:
$ {y}'(x) = \frac{(a^{2}-2(x^{2}+y^{2}))x}{(2(x^{2}+y^{2})+a^{2})y} $
Side question: Although I find this mechanically, I don't fully understand how to interpret the derivative of something that isn't a function. How does this work?
Then the question asks for the points on the original graph where the tangent is parallel to the x-axis. The solutions give $x = \pm\frac{1}{4}a\sqrt{6}$ and $y = \pm\frac{1}{4}a\sqrt{2}$, which I haven't been able to find (after setting the above derivative to $0$). Can anyone help me with this question? Thank you in advance!
Assuming your derivative is correct, it needs to be 0 so first you get $a^2/2 = x^2 + y^2$. Then use that to replace $x^2 + y^2$ in the curve equation which gives $a^2/4 = x^2 - y^2$. Then add those two equations together and you can get x in terms of $a$ and you are virtually home free.