Vertical Tangents?

The tangent line to a curve is vertical when the slope is undefined, i.e. when the derivative of the curve does not exist.

Taking the derivative of $xy^2-x^3y=6$ by implicit differentiation gives $(x*2y\dfrac{dy}{dx} + 1*y^2)-(x^3*\dfrac{dy}{dx} +3x^2*y) \Rightarrow \dfrac{dy}{dx}$ $=\large{\frac{3x^2y-y^2}{2xy-x^3}}$ , which will be undefined when $2xy-x^3=0$ .

Factoring $2xy-x^3=x(2y-x^2)$ makes the trivial solution $x=0$ to this equation immediately apparent, but since the original curve itself does not exist at $x=0$ , this answer is extraneous . Instead, rather, solve for $2y-x^2=0 \rightarrow y=\large{\frac{x^2}{2}}$ by plugging back into the original curve:

$xy^2-x^3y=6 \rightarrow x*(\large{\frac{x^2}{2})^2}$ $-(x^3*\large{\frac{x^2}{2}}$ $))$ = $6 \rightarrow x^5/4-x^5/2=6 \rightarrow -x^5/4 = 6 \rightarrow -x^5=24 \rightarrow x =$ $\boxed{\sqrt[5]{-24}}$ , which is a point on the original curve.