We know from multivariable calculus that if $y(x)$ is a function given implicitly by the equation $F(x,y) = 0$, then $$ \frac{dy}{dx} = -\frac{F_x}{F_y} \tag{1} $$ This is quickly proved by applying the multivariable chain rule to $\frac{d}{dx}F(x,y(x))=0$.
There is also a formula for the second derivative of $y$, but it is more complicated: $$\frac{d^2 y}{dx^2} = -\frac{F_{xx}F^2_y - 2F_{xy}F_xF_y+F_{yy}F^2_x}{F^3_y} \tag{2}$$ How is the formula (2) derived?
I think it's pretty clear, despite the question in Tanner Swett's comment, that $F(x, y)$ is a sufficiently smooth function of the two variables $x$ and $y$ that the equation $F(x, y) = 0$ defines $y(x)$ as an implicit function of $x$; that is, $F(x, y(x)) = 0$. Of course lurking behind such a definition of $y(x)$ is the implicit function theorem and the hypothesis that $F_y \ne 0$. Under such circumstances we have, for $F(x, y) = c$, $c$ a constant, that
$F_x + F_y y'(x) = dF(x, y(x)) / dx = 0, \tag{1}$
so that
$y'(x) = -F_x / F_y; \tag{2}$
then
$\frac{d^2y}{dx^2} = y''(x) = \frac{d}{dx}(-F_x / F_y) = -\frac{d}{dx}(F_x / F_y). \tag{3}$
We compute, using the standard formula for $(f/g)'$, $(f/g)' = (f'g - fg') / f^2$, and the fact that $\frac{d}{dx}G(x, y(x)) = G_x + G_yy'$ for any sufficiently differentiable function of two variables $G(x, y)$:
$-\frac{d}{dx}(F_x / F_y) = -((F_{xx} + F_{xy}y')F_y - F_x(F_{yx} + F_{yy}y'))/F^2_y, \tag{4}$
and into this we substitute $-F_x / F_y$ for $y'$, and then grind by diligently turning the drive crank of the algebra machine:
$((F_{xx} + F_{xy}y')F_y - F_x(F_{yx} + F_{yy}y'))/F^2_y$ $ = ((F_{xx} + F_{xy}(-F_x / F_y))F_y - F_x(F_{yx} + F_{yy}(-F_x / F_y)))/F^2_y$ $=(F_{xx}F_y - F_{xy}F_x - F_{yx}F_x + F_{yy}F_x^2/F_y)/F^2_y. \tag{5}$
Now use the "sufficiently differentiable" clause to conclude $F_{xy} = F_{yx}$ and then simply multiply the numerator and denominator by $F_y$, and turn the crank on that mill just one more time; the refined product is
$y''(x) = -\frac{d}{dx}(F_x / F_y) = -(F_{xx}F_y^2 - 2F_{xy}F_xF_y+ F_{yy}F_x^2)/F^3_y, \tag{6}$
as per request.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!