<s>[INST] Implicitly find the second derivative Given the formula $x^{2}y^{2}-8x=3$, find the second derivative.
I calculated the first derivative as $$-\frac{xy^{2}+4}{x^{2}y}$$
Working from that, I calculated the second derivative starting with
$$\frac{([x^{2}y)\frac{d}{dx}(-xy^{2}+4)]-[(-xy^{2}+4)\frac{d}{dx}(x^{2}y)]}{(x^{2}y)^{2}}$$
The left $\frac{d}{dx}$ was calculated by
$$\frac{d}{dx}(-xy^{2}+4)= [\frac{d}{dx}(-xy^{2})]+[\frac{d}{dx}(4)]= -x\frac{d}{dx}(y^{2})+ y^{2}\frac{d}{dx}(-x)= -2xy\frac{dy}{dx}- y^{2}$$
The right $\frac{d}{dx}$ was calculated by
$$\frac{d}{dx}(x^{2}y)= x^{2}\frac{d}{dx}(y)+ y\frac{d}{dx}(x^{2})= x^{2}\frac{dy}{dx}+2xy$$
Plugging everything into the formula resulted in
$$\frac{[(x^{2}y)(-2xy\frac{d}{dx}-y^{2}] - [(-xy^{2}+4)(x^2\frac{dy}{dx}+2xy)]}{(x^{2}y)^2}$$
$$\frac{[(-2x^{2}y^{2}\frac{dy}{dx}- 2x^{2}y^{3}]- [(x^{3}y^{2}\frac{dy}{dx}+2x^{2}y^{3}- 4x^2\frac{dy}{dx}-8xy]}{x^{4}y{2}}$$
Combining like terms resulted in
$$\frac{-x^{3}y^{2}\frac{dy}{dx}-4x^{2}\frac{dy}{dx}-8xy}{x^{4}y^{3}}$$
This is where I stall out.  All my previous examples haven't been in quotient form, and how do I produce a $\frac{dy}{dx}$ from a quotient?
 [/INST] It's actually a bit simpler to work directly with the original (since then you don't have to worry about the quotient rule). Simply take derivatives twice, and then solve for $y''$ in terms of $x$, $y$, and $y'$; only then plug in $y'$.
Start with
$$x^2y^2 - 8x=3.$$
Taking derivatives once, we get
\begin{align*}
\frac{d}{dx}\Bigl(x^2y^2 - 8x\Bigr) &= \frac{d}{dx}3\\
x^2\frac{d}{dx}y^2 + y^2\frac{d}{dx}x^2 - 8 &= 0\\
x^2\bigl(2yy'\bigr) + y^2\bigl(2x\bigr) -8 &= 0\\
2x^2yy' + 2xy^2 - 8&= 0\\
x^2yy' + xy^2 - 4&= 0.
\end{align*}
Now take derivatives again, and solve for $y''$:
\begin{align*}
\frac{d}{dx}\Bigl(x^2yy' + xy^2-4\Bigr) &= 0\\
x^2y\left(\frac{d}{dx}y'\right) + x^2y'\left(\frac{d}{dx}y\right) + yy'\left(\frac{d}{dx}x^2\right)\\
\quad+x\left(\frac{d}{dx}y^2\right) + y^2\left(\frac{d}{dx}x\right) &=0\\
x^2yy'' + x^2(y')^2 + 2xyy' + 2xyy' + y^2 &=0\\
x^2yy'' +x^2(y')^2 + 4xyy' + y^2 &=0\\
x^2yy'' &= -\Bigl( x^2(y')^2 + 4xyy' + y^2\Bigr)\\
y''&= -\frac{x^2(y')^2 + 4xyy' + y^2}{x^2y}\\
y''&= -\frac{(y')^2}{y} - \frac{4y'}{x} - \frac{y}{x^2}.
\end{align*}
If you want to get the value entirely in terms of $x$ and $y$, you can go back to the formula we had with the first derivative,
$$x^2yy' + xy^2 - 4 = 0,$$
we can solve for $y'$ to get
\begin{align*}
x^2yy' &= 4 - xy^2\\
y' &= \frac{4-xy^2}{x^2y}\\
y' &= \frac{4}{x^2y} - \frac{y}{x}.
\end{align*}
So we can plug in this value of $y'$ into the formula for $y''$:
\begin{align*}
y'' &= -\frac{(y')^2}{y} - \frac{4y'}{x} - \frac{y}{x^2}\\
y'' &= -\frac{1}{y}\left(\frac{4}{x^2y} - \frac{y}{x}\right)^2 - \frac{4}{x}\left(\frac{4}{x^2y}-\frac{y}{x}\right) - \frac{y}{x^2}\\
y'' &= -\frac{1}{y}\left(\frac{16}{x^4y^2} - \frac{8y}{x^3y} + \frac{y^2}{x^2}\right)-\frac{16}{x^3y} + \frac{4y}{x^2} - \frac{y}{x^2}\\
y'' &= -\frac{16}{x^4y^3} + \frac{8}{x^3y} - \frac{y}{x^2} - \frac{16}{x^3y} + \frac{3y}{x^2}\\
y'' &= -\frac{16}{x^4y^3} - \frac{8}{x^3y}+ \frac{2y}{x^2}.
\end{align*}
</s>