Answer
$$
y^{2}=Cx^{3},
$$
the orthogonal trajectories of the given family are
$$
3y^{2}+ 2x^{2}=K
$$
where $K$ is an arbitrary constant.
Work Step by Step
$$
y^{2}=Cx^{3},
$$
First, solve the given equation by differentiating implicitly with respect to $x$ we obtain the differential equation:
$$
2y y^{\prime}=3C x^{2}
$$
$\Rightarrow $
$$
\begin{aligned} y^{\prime} &=\frac{3Cx^{2}}{2y}\\
\\ &=\frac{3x^{2}}{2y} \left( \frac{y^{2}}{x^{3}}\right) \\
& =\frac{3y}{2x}
\end{aligned}
$$
Slope of given family is
$$
y^{\prime}=\frac{3y}{2x}
$$
Because represents the slope of the given family of curves at it follows that
the orthogonal family has the negative reciprocal slope so,
$$
y^{\prime}=\frac{-2x}{3y}
$$
Now you can find the orthogonal family by separating variables and integrating.
$$
3 \int y dy = -2 \int x dx
$$
$\Rightarrow$
$$
\frac{3}{2}y^{2}=-x^{2}+K \Rightarrow 3y^{2}+ 2x^{2}=K
$$
where $K$ is an arbitrary constant.
