If the area of the figure bounded by a curve, the x-axis, and two ordinates, one of which is constant, the other variable is equal to the ratio of the cube of the variable ordinate to the variable abscissa, then the curve is 

According to the given condition 

${\int }_0^x f\left(t\right)dt=\frac{y^3}{x}$

Differentiating, we get 

$\begin{array}{r}y=\frac{3y^2}{x}\frac{dy}{dx}-\frac{y^3}{x^2}\\ \frac{x^{2}y+y^3}{x^2}=\frac{3y^2}{x}\frac{dy}{dx}\\ \Rightarrow \frac{x^{2}y+y^3}{3y^{2}x}=\frac{dy}{dx}\end{array}$

Putting $y=Vx$

$\begin{array}{r}\frac{V+V^3}{3V^2}=V+x\frac{dV}{dx}\\ \Rightarrow \frac{1+V^2}{3V}-V=x\frac{dV}{dx}\\ \Rightarrow \frac{1-2V^2}{3V}=x\frac{dV}{dx}\\ \Rightarrow \frac{3V}{1-2V^2}dV=\frac{dx}{x}\end{array}$

Integrating, we have

$\begin{array}{l}\Rightarrow \frac{-3}{4}\log \left(1-2V^2\right)=\log x+\text{ Const }\\ \Rightarrow x^4{\left(1-2V^2\right)}^3=\text{ Const }\\ \Rightarrow {\left(2y^2-x^2\right)}^3=C{x}^2\end{array}$