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
 A: Correcting your error in computing $f_x$ the problem amounts to solution of the equation system:
$$
\begin{array}{rcl}
x^3+y^2&=3\\
3 \lambda  x^2+2 y&=0 \\
2x+2 \lambda  y&=2 \\
\end{array}
$$
Multiplying the second equation by $\lambda$ and subtracting the third one we obtain:
$$
\lambda^2=\frac{2(x-1)}{3x^2}.
$$
Substituting the obtained value into the second equation one obtains:
$$4y^2=9\lambda^2x^4\implies y^2=\frac32x^2(x-1),$$
which after substitution into the first equation gives rise to:
$$
5x^3-3x^2-6=0.
$$
The equation has one real 
$$x_r=\frac{1+\omega+\omega^{-1}}5, \text{ with } 
\omega=\left(76+5\sqrt{231}\right)^{1/3}
$$
and two complex roots. From the context of your question the complex roots can be disregarded. Since the real root is larger than $1$ the pair 
$$(x,y)=\left(x_r,x_r\sqrt{\frac{3(x_r-1)}2}\right)\approx(1.304821,0.882306)$$ 
is the only candidate for the solution (you should however check if the stationary point is indeed the global maximum).
A: 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.
