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.

First, transform your equation:
$$b^2x^2+a^2y^2=a^2b^2$$
$$a^2y^2=a^2b^2-b^2x^2$$
$$y^2=b^2-\frac{b^2}{a^2}x^2$$
$$xy^2=b^2x-\frac{b^2}{a^2}x^3$$
Next, differentiate and solve to find extrema:
$$b^2-3\frac{b^2}{a^2}x^2=0$$
$$b^2=3\frac{b^2}{a^2}x^2$$
$$a^2=3x^2$$
$$x=\pm\sqrt{\frac{1}{3}}a$$
Finally, evaluate $xy^2$ at $x=+\sqrt{\frac{1}{3}}a$:
$$xy^2=b^2x-\frac{b^2}{a^2}x^3=b^2\frac{a}{\sqrt{3}}-\frac{b^2}{a^2}\frac{a^3}{3\sqrt{3}}=\frac{ab^2}{\sqrt{3}}-\frac{ab^2}{3\sqrt{3}}=\frac{2}{3\sqrt{3}}ab^2$$
That's your maximum. The minimum is at $x=-\sqrt{\frac{1}{3}}a$:
$$xy^2=b^2x-\frac{b^2}{a^2}x^3=-b^2\frac{a}{\sqrt{3}}+\frac{b^2}{a^2}\frac{a^3}{3\sqrt{3}}=-\frac{ab^2}{\sqrt{3}}+\frac{ab^2}{3\sqrt{3}}=-\frac{2}{3\sqrt{3}}ab^2$$
Wolfram Alpha seems to agree.
P.S. Let's also solve the original equation for $y$ at $x=\pm\sqrt{\frac{1}{3}}a$:
$$b^2x^2+a^2y^2=a^2b^2$$
$$b^2\frac{a^2}{3}+a^2y^2=a^2b^2$$
$$\frac{b^2}{3}+y^2=b^2$$
$$y=\pm\sqrt{\frac{2}{3}}b$$
