The equation of the tangent line at the point (1,-1) is $y=\frac{4x}{3}-\frac{7}{3}$. Given the equation, $x^2y+ay^2=b$, find the values of a and b. So I got this question:
"The equation of the tangent line at the point $(1,-1)$ is $y= \dfrac{4x}{3} - \dfrac{7}{3}$. Given the equation, $x^2y + ay^2 = b$, find the values of $a$ and $b$."
I'm really stuck right now because I think I am supposed to differentiate the equation with $a$ and $b$ in terms of $x$, but when I do I'm stuck with $\dfrac{\mathrm dy}{\mathrm dx}$ on one side and $\dfrac{\mathrm da}{\mathrm dx}$ on the other. So even if I subbed in the $(1, -1)$ and used algebra to get $\dfrac{\mathrm dy}{\mathrm dx}$ to equal $\dfrac{4}{3}$ (slope I want to get), I am stuck with $\dfrac{\mathrm dy}{\mathrm dx}$ AND a $\dfrac{\mathrm da}{\mathrm dx}$ term I don't know how to get rid of. Thanks in advance for the help!

Hint
As $(1,-1)$ lies on the given curve,  $$1+a=b$$
As $a,b$ are conventionally considered arbitrary constants,
using https://en.m.wikipedia.org/wiki/Chain_rule
$$\dfrac{d(x^2)}{ydx}+\dfrac{x^2d(y^{-1})}{dy}\cdot\dfrac{dy}{dx}+2a\dfrac{d(y^{-2})}{dx}=0$$
$$\dfrac{2x}y-\dfrac{x^2}{y^2}\cdot\dfrac{dy}{dx}-\dfrac{2a}{y^3}\dfrac{dy}{dx}=0$$
