Answer
$$
y=Ce^{x},
$$
the orthogonal trajectories of the given family are
$$
y^{2}+ 2x=K
$$
where $K$ is an arbitrary constant.
Work Step by Step
$$
y=Ce^{x},
$$
First, solve the given equation by differentiating implicitly with respect to $x$ we obtain the differential equation:
$$
y^{\prime}=Ce^{x}= y
$$
Slope of given family is
$$
y^{\prime}=y
$$
Because represents the slope of the given family of curves at it follows that
the orthogonal family has the negative reciprocal slope so,
$$
y^{\prime}=\frac{-1}{y}
$$
Now you can find the orthogonal family by separating variables and integrating.
$$
\int y dy = - \int dx
$$
$\Rightarrow$
$$
\frac{1}{2}y^{2}=-x+K \Rightarrow y^{2}+ 2x=K
$$
where $K$ is an arbitrary constant.