Assume the velocity field: $$ v_x=-Ue^{-at}\cos \left( \frac{\pi x}{L} \right)\sin \left( \frac{\pi y}{L} \right) $$ $$ v_y=-Ue^{-at}\sin \left( \frac{\pi x}{L} \right)\cos \left( \frac{\pi y}{L} \right) $$
The streamline equation is the solution of the differential equation: $$ ds=\frac{dx}{v_x}=\frac{dy}{v_y} $$ $$ \iff -Ue^{-at}\sin \left( \frac{\pi x}{L} \right)\cos \left( \frac{\pi y}{L} \right) dx=-Ue^{-at}\sin \left( \frac{\pi x}{L} \right)\cos \left( \frac{\pi y}{L} \right)dy $$ $$ \Rightarrow \tan \left( \frac{\pi x}{L} \right)dx=\tan \left( \frac{\pi y}{L} \right)dy $$
indefinite integration and adjusting of the constant gives: $$ \left| \cos \left( \frac{\pi x}{L} \right)\right|=c \cdot \left|\cos \left( \frac{\pi y}{L} \right)\right| $$ $$ \iff \cos^2 \left( \frac{\pi x}{L} \right)+\lambda \cdot \cos^2 \left( \frac{\pi y}{L} \right)=0 $$
Given that we are given no initial point $(x_0,y_0)$ to determine $c$, how can a final result be obtained?
You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes
\begin{equation} \int_{x_0}^x \tan \left( \frac{\pi x}{L} \right) \mathrm{d} x = \int_{y_0}^y \tan \left( \frac{\pi y}{L} \right) \mathrm{d} y \end{equation}
which gives
\begin{equation} - \frac{L}{\pi} \left[ \ln \left| \cos( \frac{\pi x}{L} ) \right| \right]_{x_0}^x = - \frac{L}{\pi} \left[ \ln \left| \cos( \frac{\pi y}{L} ) \right| \right]_{y_0}^y \end{equation}
or
\begin{equation} \ln\left| \frac{\cos( \frac{\pi x}{L} )}{\cos( \frac{\pi x_0}{L} )} \right| = \ln\left| \frac{\cos( \frac{\pi y}{L} )}{\cos( \frac{\pi y_0}{L} )} \right| \end{equation}
Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is
\begin{equation} f(x,y) = \left( \frac{ \cos( \frac{\pi x}{L} )}{ \cos( \frac{\pi y}{L} )} \right)^2 - \underbrace{ \left( \frac{\cos( \frac{\pi x_0}{L} ) }{\cos( \frac{\pi y_0}{L} )}\right)^2}_{c^2} = 0. \end{equation}