The slope of the hyperbola $b^2 x^2 - a^2y^2 = a^2 b^2$ at the upper end of its right-hand latus rectum is $4/3$. What is the eccentricity? How to approach this type of problem?

The slope of the curve $b^2 x^2 - a^2y^2 = a^2 b^2$ at the upper end of its latus rectum to the right of the origin is $4/3$. What is the eccentricity of the curve? 

I get the equation of derivative of $y'= \dfrac{xb^2}{ya^2}$. I still don't have a point of $x$ and $y$ for latus rectum to insert in equation $e=\sqrt {a^2 + b^2}/a$

As mentioned by @Blue, the latus rectum is a vertical line through the focus $(c,0) \equiv (ae,0)$.
The abcissa of the latus rectum is $ae = \sqrt{a^2+b^2}.$ So, to find the $y$ coordinates of its terminii, we have $$b^2a^2e^2 - a^2y^2 = a^2b^2$$ $$\implies y^2 = b^2(e^2-1)$$
$$\implies y' = \frac 43 = {aeb^2\over b\sqrt{e^2-1} a^2} =\frac ba\cdot{e\over\sqrt{e^2-1}}$$
Note that $$e^2 = 1+{b^2\over a^2}$$
$$\implies \sqrt{e^2-1} = \frac ba$$
$$\implies \boxed{e = \frac 43}$$
