Maximize $(x-1)2y$ subject to $x^3+y^2=3$ I have to do with lagrange multiplier
f = $(x-1)2y + \lambda (x^3+y^2-3) = 0 $
I get 
$f_x= y + \lambda x = 0 $
$f_y = y\lambda + x - 1 = 0 $ 
$x^3+y^2=3$
and 
how to solve this

Hint.
Solving for $x, y$
$$
\left\{
\begin{array}{rcl}
 3 \lambda  x^2+2 y&=&0 \\
 2 (x-1)+2 \lambda  y&=&0 \\
\end{array}
\right.
$$
we get
$$
x = \frac{1\pm\sqrt{1-6 \lambda ^2}}{3 \lambda ^2}\\
y = \frac{3\pm\frac{\sqrt{1-6 \lambda ^2}}{\lambda^2}-\frac{1}{\lambda ^2}}{3 \lambda }
$$
substituting those values into $x^3+y^2 = 3$
$$
\left\{
\begin{array}{c}
 \frac{\left(1-\sqrt{1-6 \lambda ^2}\right)^3}{27 \lambda ^6}+\frac{\left(\frac{\sqrt{1-6 \lambda ^2}}{\lambda
   ^2}-\frac{1}{\lambda ^2}+3\right)^2}{9 \lambda ^2}-3=0 \\
 \frac{\left(\sqrt{1-6 \lambda ^2}+1\right)^3}{27 \lambda ^6}+\frac{\left(-\frac{\sqrt{1-6 \lambda ^2}}{\lambda
   ^2}-\frac{1}{\lambda ^2}+3\right)^2}{9 \lambda ^2}-3=0 \\
\end{array}
\right.
$$
after symplifying and solving we obtain
$$
\lambda^* = \pm 0.34548248
$$
so we get
$$
\left[
\begin{array}{cc}
x^*& y^*\\
 1.30482 & -0.882306 \\
 1.30482 & 0.882306 \\
 4.28061 & -9.49573 \\
 4.28061 & 9.49573 \\
\end{array}
\right]
$$
as the stationary points.
