Similar to equipotential?

Since $u(0,y) = e^{-y^2}$ , we can assume that $u(x,y) = e^{v(x)-y^2}$ . Then we have:

$\begin{aligned} y\frac {\partial u}{\partial x} + x \frac {\partial u}{\partial y} & = 0 \\ y\frac {dv}{dx} e^{v(x)-y^2} - 2xye^{v(x)-y^2} & = 0 \\ \implies \frac {dv}{dx} & = 2x \\ v(x) & = \int 2x \ dx = x^2 + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ v(0) & = 0+C = 0 & \small \color{#3D99F6} \implies C = 0 \\ \implies v(x) & = x^2 \\ \implies u(x,y) & = e^{x^2-y^2} \\ u(1,2) & = e^{1^2-2^2} \\ & = e^{-3} \approx \boxed{0.0498} \end{aligned}$