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!
 A: First, you have to differentiate both sides implicitly:
$$\frac{\mathrm d}{\mathrm dx} \left(x^2y+ay^2\right) = \frac{\mathrm d}{\mathrm dx} (b)$$
$x^2y$ is a product, so $\dfrac{\mathrm d}{\mathrm dx} \left(x^2y\right) = y\dfrac{\mathrm d}{\mathrm dx} \left(x^2\right)+x^2\dfrac{\mathrm dy}{\mathrm dx} = 2xy+x^2\dfrac{\mathrm dy}{\mathrm dx}$.
$a$ is a constant, so $\dfrac{\mathrm d}{\mathrm dx}\left(ay^2\right) = a\dfrac{\mathrm d}{\mathrm dx}\left(y^2\right) = 2ay\dfrac{\mathrm dy}{\mathrm dx}$.
$b$ is also a constant, so $\dfrac{\mathrm d}{\mathrm dx} (b) = 0$.
Combining these gives:
$$2xy+x^2\frac{\mathrm dy}{\mathrm dx}+2ay\frac{\mathrm dy}{\mathrm dx} = 0 \tag{1}$$
Notice how there's no $\dfrac{\mathrm da}{\mathrm dx}$ anywhere. I would guess the issue here is that you're differentiating $ay^2$ the same way you'd differentiate $xy^2$ by keeping an extra $\dfrac{\mathrm da}{\mathrm dx}$. Even if you did use the product rule, since $\color{blue}{\dfrac{\mathrm da}{\mathrm dx} = 0}$, you'd get the same result:
$$\dfrac{\mathrm d}{\mathrm dx} \left(ay^2\right) = y^2\dfrac{\mathrm d}{\mathrm dx} (a)+a\dfrac{\mathrm d}{\mathrm dx}\left(y^2\right) = y^2\color{blue}{\dfrac{\mathrm da}{\mathrm dx}}+2ay\dfrac{\mathrm dy}{\mathrm dx} = 2ay\dfrac{\mathrm dy}{\mathrm dx}$$
Finally, the equation of the tangent line at $(1, -1)$ is $y = \dfrac{4}{3}x-\dfrac{7}{3}$, so the slope becomes $\dfrac{\mathrm dy}{\mathrm dx}\biggr\rvert_{x = 1} = \dfrac{4}{3}$.
If you use the point $(1, -1)$ for equation $(1)$, the only unknown left is $a$, which you can solve for.
Finally, since the tangent and the original curve share the point $(1, -1)$, plugging in $x = 1$, $y = -1$, and the value of $a$ (from the previous part) will give $b$.
A: 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
$$\dfrac{d(x^2)}{ydx}+\dfrac{x^2d(y^{-1})}{dy}\cdot\dfrac{dy}{dx}+2a\dfrac{d(y^{-2})}{dx}=0$$
$$\dfrac{2x}y-\dfrac{x^2}{y^2}\cdot\dfrac{dy}{dx}-\dfrac{2a}{y^3}\dfrac{dy}{dx}=0$$
