One of our final exam exercise sheets features this particular exercise :
Find the extrema of : $2xy^3+y-x^2=0$, where $y=y(x)$ . As I thought it, this exercise involves the implicit function theorem, so I defined $F(x,y(x))=2xy^3+y-x^2=0$.
Then I computed $\frac{\partial F}{\partial x}=y^3+3xy^2y'-x=0 $ and $\frac{\partial F}{\partial y}=6xy^2+1=0 $ and got $y'=\frac{x-y^3}{3y^2x}$ .
But I'm not really sure on how to go about finding the extrema of $y(x)$ .. Any ideas ?
$$ \begin{equation*} F(x,y(x)) = 2xy^3 + y - x^2 = 0 \\ \implies F_x = 2y^3 + 6xy^2y' + y' - 2x = 0 \\ \implies 2x-2y^3 = y'(6xy^2 +1) \\ \implies y' = \frac{2(x-y^3)}{6xy^2 + 1}, \end{equation*}$$
using the product rule and chain rule on the $2xy^3$ term.
Critical points at $y' = 0$, so
$$ \begin{align*} \frac{2(x-y^3)}{6xy^2 + 1} &=0 \\ \implies x-y^3 &= 0 \\ \implies y^3 &= x. \end{align*} $$
Sub $y^3$ in place of $x$ into the original equation $F(x,(y(x))=0$ and solve for $y$.
I get critical points at $\left( 0, 0\right)$ and $\left( -1, -1 \right).$
Hope that helps. :)
Edit: Now that I look at it, I'm sure you're doing the partial differentiation wrong. Bear in mind that $y$ is a function of $x$, so you should be using product and chain rule as I did. After that it's pretty much just algebra. Also, my use of $F_x$ is just an alternative notation for $ \frac{\partial F}{\partial x}$. :)