I'm looking at this question
If the tangent at the point $P$ with coordinates $(h, k)$ on the curve $y^2 = 2x^3$ is perpendicular to the line $4x = 3y$, find $(h, k).$
This is how I attempted it
$\frac{d}{dx}(y^2) = 2\frac{d}{dx}x^3$
$2y \frac{dy}{dx} = 6x^2$
$\frac{dy}{dx} = \frac{3x^2}{y}$
Since the tangent is perpendicular to the line $4x = 3y$, it's slope must be $\frac{-3}{4}$.
Therefore we have
$\frac{3x^2}{y} = \frac{-3}{4}$ => $y = -4x^2$
Squaring and dividing the equation of the curve with this, we get two solutions $(h, k) = (\frac{1}{8}, \frac{-1}{16})$ and $(h, k) = (0, 0)$.
I happened to check the graph of $y^2 = 2x^3$ on Wolfram Alpha and found that the point $(0, 0)$ is actually a point of singularity.
I am in doubt whether $(h, k) = (0, 0)$ is a valid answer to this question.
A priori the solution set $L$ of the problem is a certain subset of ${\mathbb R}^2$, which might even be empty. Doing some calculations (which I have not checked) you have proven in a $$\ldots\Rightarrow\ldots\Rightarrow\ldots\Rightarrow\ldots$$ way that $$L\subset L':=\left\{\biggl({1\over 8}, -{1\over16}\biggr), (0,0)\right\}\ .$$
Now if you had a general theory guaranteeing you exactly $2$ solutions to this problem, you would be done: You could confidentially say that $L=L'$. But there is no such theory. Therefore it remains to check which of the points in $L'$ are indeed solutions of the original problem. This seems to be the case for $\bigl({1\over 8}, -{1\over16}\bigr)$, but is certainly not the case for $(0,0)$.
The curve $y^2=2x^3$, or $$y=\pm {1\over\sqrt{2}} \,x^{3/2}=: \pm\, f(x)\qquad (x\geq0)\ ,$$ has a cusp singularity (called a "Rückkehrpunkt" in German) at $(0,0)$ . At any rate, from above and below the limiting tangent direction at $(0,0)$ is horizontal, since $$\lim_{x\to 0+} f'(x)=\lim_{x\to0+}{3\over 2\sqrt{2}}x^{1/2}=0\ ,$$ and does not have slope $-{3\over4}$, as required by the problem.