Calculating tangent at two points $(4a,8a)$ Find the equation of the tangent to the curve $ay^2=x^3$ at the points $(4a,8a)$
I have re-arranged the equation to get
$$y = \left(\frac{x^3}{a}\right)^{\frac{1}{2}}$$
Then taking its derivative  I get
$$\frac{dy}{dx}=\frac{3\left(\sqrt{\frac{x^3}{a}}\right)}{2x}$$,
when $x=4a$, I find that $\frac{dy}{dx}=3$
and so I have tried:
$$\frac{y-\left(\frac{x^3}{a}\right)^{\frac{1}{2}}}{x-4a}=3$$
However, this will not give me a simple answer which is to be $y = 3x-4a$
How do I correctly calculate at both points?

You have probably miscalculated the derivative when the chain rule was applied.
Note that $\dfrac{du^\frac{1}{2}}{du}=\dfrac{1}{2}u^{-\frac{1}{2}}$.
Hence,
$y'=\dfrac{d}{d\dfrac{x^3}{a}}(\dfrac{x^3}{a})^\frac{1}{2}\cdot\dfrac{d}{dx}(\dfrac{x^3}{a})$
$=\dfrac{1}{2}\cdot(\dfrac{x^3}{a})^{-\frac{1}{2}}\cdot\dfrac{3x^2}{a}$
$=\dfrac{3}{2}\sqrt{\dfrac{x}{a}}$
$y'|_{x=4a}=3$
Therefore, the equation of tangent:
$\dfrac{y-8a}{x-4a}=3$
$\underline{\underline{y=3x-4a}}$
