Answer
\[3y^2+x^2=K\]
Work Step by Step
$y=cx^3$ ____(1)
Differentiate (1) with respect to $x$
\[\frac{dy}{dx}=3cx^2\]
From (1)
$\large{\frac{dy}{dx}=3\left(\frac{y}{x^3}\right)x^2=\frac{3y}{x}}$___(2)
Replace $\Large\frac{dy}{dx}$ by $\Large-\frac{dx}{dy}$ in (2) [ For orthogonal Trajectories]
\[-\frac{dx}{dy}=\frac{3y}{x}\]
Separating variables
\[-x dx=3y dy\]
Integrating, $$C_{1}-\int x dx=3\int y dy$$
$C_{1}$ is constant of jntegration
\[C_{1}-\frac{x^2}{2}=\frac{3y^2}{2}\]
$3y^2+x^2=K$, where $K=2C_{1}$
Hence orthogonal trajectories of (1) is $3y^2+x^2=K$
