<s>[INST] Finding the Maximum value. 
Maximize $xy^2$ on the ellipse $b^2x^2 +a^2y^2= a^2b^2$

The steps I tried to solve:
$$\nabla f = (y^2,2yx)\lambda\qquad g = (2xb^2,2y^2a^2)\lambda$$
$$y^2= 2xb^2\lambda$$
$$2yx= 2y^2a^2\lambda$$

$$
\left.
\begin{array}{l}
\text{}&y^2= 2xb^2\lambda\\
\text{}&
\end{array}
\right\}
*a^2y
$$    
$$
\left.
\begin{array}{l}
\text{}&2yx= 2y^2a^2\lambda\\
\text{}&
\end{array}
\right\}
*b^2x
$$ 

$$y^3a^2= 2yxa^2b^2 \lambda$$      
$$2yx^2b^2 = 2y^2a^2b^2x\lambda$$
$\color{maroon}{\mathbf{Equalize}}$

$$2a^2b^2xy = 2a^2b^2xy^2$$
$$y=1$$
My main problem is which equations does one set each equal to. If anyone knows which one equals the other this will lead me to the right place in finding the solution.
 [/INST] Since you started off using Langrange multipliers, let’s continue down that path. Using the correct value for $g_y$, we have $$\begin{align}
y^2 &= 2b^2\lambda x \\
2xy &= 2a^2\lambda y
\end{align}$$ which upon eliminating $\lambda$ gives $$
a^2y^3 = 2b^2x^2y,
$$ so either $y=0$ or $a^2y^2=2b^2x^2$. Substituting this into the constraint: $$b^2x^2+2b^2x^2=a^2b^2 \\
\text{or} \\
x^2=\frac13a^2.$$ Plugging this back into the constraint and solving for $y$: $$\frac13a^2b^2+a^2y^2=a^2b^2 \\
\text{or} \\
y^2=\frac23b^2.$$ $f$ is an even function of $y$, so it doesn’t matter which root we use to test these points against $f$. $$
f\left(+\sqrt{\frac13}a,\pm\sqrt{\frac23}b\right) = {2\sqrt3 \over 9}ab^2 \\
\text{and} \\
f\left(-\sqrt{\frac13}a,\pm\sqrt{\frac23}b\right) = -{2\sqrt3 \over 9}ab^2
$$ We still have $y=0$ as a candidate, but $f(x,0)=0$, so this doesn’t yield an extremum.
</s>