\begin{cases} xu_x+yu_y+zu_z=4u\\ u(x,y,1)=xy\\ \end{cases}
Using Lagrange method we get:
$$\frac{x}{y}=c_1, \frac{y}{z}=c_2, \frac{z}{\sqrt[4]{u}}=c_3$$
So the general solution is $$u=\frac{z^4}{c_3}$$?
\begin{cases} xu_x+yu_y+zu_z=4u\\ u(x,y,1)=xy\\ \end{cases}
Using Lagrange method we get:
$$\frac{x}{y}=c_1, \frac{y}{z}=c_2, \frac{z}{\sqrt[4]{u}}=c_3$$
So the general solution is $$u=\frac{z^4}{c_3}$$?
Converting to spherical coordinates, we get
$$ru_r = 4u \implies u = f(\theta,\phi)r^4$$
Then plugging in our boundary condition at $r\cos\theta = 1$ and $xy=r^2\sin^2\theta\sin\phi\cos\phi$, we can get
$$f(\theta,\phi)r^4 = r^2\sin^2\theta\sin\phi\cos\phi\cdot(1)=r^2\sin^2\theta\sin\phi\cos\phi\cdot (r^2\cos^2\theta)$$
$$\implies f(\theta,\phi) = \cos^2\theta\sin^2\theta\sin\phi\cos\phi$$
by canceling out the $r^4$ on both sides. In other words when we convert back to Cartesian we get the solution
$$u(x,y,z) = xyz^2$$