$$\begin{cases}-x\partial_xu +y\partial_yu =-x^2u \\ u(x,1) = e^{-x} \end{cases}$$
I recently ask a question on PDE and I hope I understand how it works. Let $z(t)=u(x(t),y(t))$
$$\begin{cases}\dot{x}=-x\\ \dot{y}=y\\ \dot{z}=-x^2z \end{cases}$$
I think I can solve either for $t$ or using the first two row eliminating the time dependence and so I get with $\alpha,\beta \in \mathbb{R}$
$$x(t)=\alpha e^{-t}\\ y(t)=\beta e^t$$
And without the time dependence - that is also the result I get from $\frac{dx}{-x}=\frac{dy}{y}$ -
$$xy=c_1\;, c_1\in \mathbb{R}$$
From the last row solving in x let $c_2 \in \mathbb{R}$
$$z=c_2\exp\bigg({-\frac{x^3}{3}}\bigg)$$
Since $c_2 = F(c_1)$
$$z=F(xy)\exp\bigg({-\frac{x^3}{3}}\bigg)\\u(1,x)=e^{-x}=F(x)\exp\bigg({-\frac{x^3}{3}}\bigg)\\ F(x) = \exp\bigg({\frac{x^3}{3}}-x\bigg) $$
And the general solution is
$$u(x,y) = \exp\bigg({\frac{(xy)^3-x^3}{3}}-xy\bigg)$$
I think this solution has sense beacause plotting $u(x,y)$ and the parametric curve $\gamma(t)=(t, \frac{1}{t}, \exp\bigg(\frac{-t^3}{3}-\frac{2}{3}\bigg))$ I get
And the red line follows the surface, I don't know if this count as a solution verification. Any clarification, correction, signaling of unclear passages is welcome thank you very much
$$-x\partial_xu +y\partial_yu =-x^2u$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{-x}=\frac{dy}{y}=\frac{du}{-x^2u}$$ A first characteristic equation comes from solving $\frac{dx}{-x}=\frac{dy}{y}$ $$xy=c_1$$ A second characteristic equation comes from solving $\frac{dx}{-x}=\frac{du}{-x^2u}$ $$e^{-x^2/2}u=c_2$$ This doesn't agree with your result.
General solution of the PDE on implicite form $c_2=F(c_1)$ : $$e^{-x^2/2}u=F(xy)$$ $F$ is an arbitrary function (to be determined according to the boundary condition). $$\boxed{u(x,y)=e^{x^2/2}F(xy)}$$
Condition :
$u(x,1)=e^{-x}=e^{x^2/2}F(x)$ $$F(x)=e^{-x-\frac12 x^2}$$ The function $F$ is determined. We put it into the above general solution where the argument is $xy$ thus $F(xy)=e^{-xy-\frac12 (xy)^2}$:
$u(x,y)=e^{x^2/2}e^{-xy-\frac12 (xy)^2}$ $$u(x,y)=e^{-xy-\frac12 (xy)^2+\frac12 x^2}$$ This is the particular solution which satisfies both the PDE and the condition.
NOTE :
Your parametrization with $t$ is equivalent to $$\frac{dx}{-x}=\frac{dy}{y}=\frac{dz}{-x^2z}=dt \quad\implies\quad \begin{cases} \frac{dx}{dt}=-x \\ \frac{dy}{dt}=y \\ \frac{dz}{dt}=-x^2z \end{cases}$$