Hello Ethan,
We are given the implicitly defined curve:
$$x^3+y^3=4xy+1$$
a.) Implicitly differentiating with respect to $x$, we obtain:
$$3x^2+3y^2\frac{dy}{dx}=4\left(x\frac{dy}{dx}+y \right)+0$$
Arranging with all terms having $$\frac{dy}{dx}$$ as a factor on the left, and everything else on the right, we have:
$$3y^2\frac{dy}{dx}-4x\frac{dy}{dx}=4y-3x^2$$
Factor the left side:
$$\left(3y^2-4x \right)\frac{dy}{dx}=4y-3x^2$$
Dividing through by $$3y^2-4x$$ we find:
$$\frac{dy}{dx}=\frac{4y-3x^2}{3y^2-4x}$$
b.) The tangents are vertical where:
$$3y^2-4x=0\implies x=\frac{3}{4}y^2$$
Substituting for $x$ into the original curve, there results:
$$\left(\frac{3}{4}y^2 \right)^3+y^3=4\left(\frac{3}{4}y^2 \right)y+1$$
$$\frac{27}{64}y^6+y^3=3y^3+1$$
Arrange in standard polynomial form:
$$27y^6-128y^3-64=0$$
Applying the quadratic formula, we find:
$$y^3=\frac{8}{27}\left(8\pm\sqrt{91} \right)$$
Hence:
$$y=\frac{2}{3}\sqrt[3]{8\pm\sqrt{91}}$$
$$x=\frac{1}{2}\sqrt[3]{155\pm16\sqrt{91}}$$
Thus, the points having vertical tangents are:
$$(x,y)=\left(\frac{1}{2}\sqrt[3]{155\pm16\sqrt{91}},\frac{2}{3}\sqrt[3]{8\pm\sqrt{91}} \right)$$
c.) $$\left.\frac{dy}{dx} \right|_{(x,y)=(2,1}=\frac{4(1)-3(2)^2}{3(1)^2-4(2)}=\frac{8}{5}$$
