Solve $3u_y+u_{xy}=0$ 
Solve $3u_y+u_{xy}=0$, does a solution exist with conditions $u(x,0)=0$ and $u_y(x,0)=0$

I made substitution $v=u_y$
To get $3v+v_x=0$
Which is solved by $v={3}e^{-3x}$
So $u_y= e^{-3x}$
then $e^{-3x}+u_{xy}=0$
$u_y=\frac{1}{3}e^{-3x}+g(y)$
$u=\frac{1}{3}e^{-3x}y+xg(y)+f(x)$
I'm not sure what to do with the initial conditions, I get $u_y=\frac{1}{3}e^{-3x}+g(0)=0$
So $g(0) = \frac{1}{3}e^{-3x}$
and $u(x,0)=e^{-3x}=0-\frac{1}{3}xe^{-3x}+f(x)$ so $f(x)=\frac{1}{3}xe^{-3x}-e^{-3x}$
But I still need to find a solution of $g(y)$to find $u$.
 A: The general solution to $v_x = -3v$ is $v(x,y) = \phi(y)e^{-3x}$ where $\phi$ some function. Thus $u_y(x,y) = \phi(y)e^{-3x}$ leads to $\Phi(y)e^{-3x} + C$ where $\Phi$ is an antiderivative of $\phi$ and $C$ is a constant.  Now,
$$u(x,0) = \Phi(0) e^{-3x} + C$$
and
$$u_y(x,0) = -3\phi(0) e^{-3x}.$$
so $u_y(x,0) = 0$ forces $\phi(0) = 0$ and $\Phi(0) e^{-3x} + C = 0$ (for all $x$) forces $\Phi(0) = 0$ and $C = 0$.
Thus any function of the form $u(x,y) = \Phi(y)e^{-3x}$ is a solution provided that $\Phi$ is differentiable, $\Phi(0) = 0$, and $\Phi'(0) = 0$.
A: Let $v = u_y$. Then:
$$3u_y+u_{xy}=0 \Rightarrow 3v + v_x = 0 \Rightarrow v_x = -3v \Rightarrow u_y = Ae^{-3x}.$$
Additionally, given an arbitrary point $(x_1, y_1)$, we have that:
$$u(x,y)-u(x_1,y_1) = \int_{y_1}^{y}u_y(x,s) ds =  \int_{y_1}^{y} A e^{-3x} ds = Ae^{-3x}(y-y_1).$$
By choosing $y_1 = 0$, we obtain:
$$\begin{cases}
u(x,y) &= Ae^{-3x}y+u(x_1,0)\\
u_y(x,y) &= Ae^{-3x}
\end{cases}.$$
Using the boundary condition $u(x,0) = 0$, we get:
$$\begin{cases}
u(x,y) &= Ae^{-3x}y\\
u_y(x,y) &= Ae^{-3x}
\end{cases}.$$
Finally, $u_y(x,0)=0$ yields to:
$$u_y(x,0) = Ae^{-3x} = 0 \Rightarrow A = 0 \Rightarrow \begin{cases}
u(x,y) &= 0\\
u_y(x,y) &=0
\end{cases}.$$
