Answer
The orthogonal trajectories of the family of curves
$$
y=e^{kx}
$$
are
$$
2 y^{2} \ln y-y^{2}=C-2 x^{2}
$$
Work Step by Step
To find the orthogonal trajectories of the family of curves
$$
y=e^{kx}
$$
First differentiate both sides:
$$
\frac{d}{d x}(y)=\frac{d}{d x}\left(e^{kx}\right) \Rightarrow y^{\prime}=k e^{x}=\frac{\ln y }{x}y,
$$
an orthogonal trajectory must have a slope that is the negative reciprocal of the tangent line slope:
$$
y^{\prime}=\frac{-x }{ y \ln y},
$$
we write the equation in terms of differentials and integrate both sides:
$$
\frac{d y}{d x}=-\frac{x}{y \ln y} \Rightarrow y \ln y d y=-x d x \Rightarrow \int y \ln y d y=-\int x d x
$$
integrate by parts with
$$
u=\ln y ,\quad\quad d v=y d y \\ du = \frac{dy}{y} ,\quad\quad v=\frac{1}{2}y ^{2}
$$
we get
$$
\frac{1}{2} y^{2} \ln y-\frac{1}{2} \int y d y= -\frac{1}{2} x^{2} +C_{1}
$$
$\Rightarrow $
$$
2 y^{2} \ln y-y^{2}=C-2 x^{2}
$$
where $C=2C_{1}$ is an arbitrary constant.
Thus the orthogonal trajectories of the family of curves
$$
y=e^{kx}
$$
are
$$
2 y^{2} \ln y-y^{2}=C-2 x^{2}
$$
