(Edit: I redesigned to a more convenient approach)
I have to find rigurously the general solution for the following partial differential equation
\begin{equation}\tag{1} u_{xx}(x,y) + y u_{x}(x,y) = 0 \quad \forall (x,y)\in \mathbb{R}^2 \end{equation}
where $u\in \mathcal{C}^2(\mathbb{R}^2)$. Here, $u_{xx},u_{xy},u_{yy}$ are continuous.
I have written the following:
Let $u$ be a solution of $(1)$. We note that $u\in C^2(\mathbb{R}^2)$. Then,
$$u_{xx}+yu_x=0 \iff e^{xy}u_{xx} + y e^{xy} u_x=0 \iff(e^{xy}u_x)_x=0$$
Thus, exists $f:\mathbb{R}\longrightarrow \mathbb{R}$ such that $e^{xy}u_x=f(y)$ which is rewritten as $u_x=f(y)e^{-xy}$.
Now, if $y=0$, exists $A\in \mathbb{R}$ such that
$$u(x,0)=f(0)x+A$$
If $y \neq 0$ we have that exists $g:\mathbb{R}\setminus \{0\} \longrightarrow \mathbb{R}$ such that
$$u(x,y)=-\frac{f(y)}{y}e^{-xy} + g(y)$$
So
$$u(x,y)=\begin{cases}-\frac{f(y)}{y}e^{-xy} + g(y), \quad &y\neq 0 \\ f(0)x+A, \quad &y=0\end{cases}$$
For $y=0$
$$ u_x(x,0)=f(0), \quad u_{xx}(x,0)=0 $$
For $y\neq 0$
$$ u_x(x,y)=f(y)e^{-xy}, \quad u_{xx}(x,y)=-yf(y)e^{-xy} $$
Now, $f(y)=u_x(0,y)$ and $f(0)=u_x(0,0)$ so $f\in C(\mathbb{R})$ and $f\in C^1(\mathbb{R}\setminus \{0\})$. For $y\neq 0$,
$$ g(y)=u(x,y)+\frac{e^{-xy}}{y}f(y) \in C^1(\mathbb{R}\setminus \{0\}) $$
So
$$ u_y(x,y)=y^{-2}e^{-xy}((1+xy)f(y)-yf'(y))+g'(y) \implies $$
$$ f'(y)=\frac{y^2u_y(1,y)-y^2u_y(0,y)-yf(y)}{y(1-e^{-y})} \in C^1(\mathbb{R}\setminus \{0\}) $$
and
$$ g'(y)=u_y(0,y)-\frac{1}{y^2}(f(y)-yf'(y)) \in C^1(\mathbb{R}\setminus \{0\}) $$
So far we have $f \in C(\mathbb{R})$, $f,g\in C^1(\mathbb{R})$. Now, because $u\in C^2 (\mathbb{R}^2)$ we have that for all $x\in \mathbb{R}$
\begin{align} &\lim_{y\to 0} u(x,y)=u(x,0), \quad \lim_{y\to 0} u_x(x,y)=u_x(x,0),\\ &\lim_{y\to 0} u_y(x,y)=u_y(x,0), \quad \lim_{y\to 0} u_{xx}(x,y)=u_{xx}(x,0),\\ &\lim_{y\to 0} u_{yy}(x,y)=u_{yy}(x,0), \quad \lim_{y\to 0} u_{xy}(x,y)=u_{xy}(x,0) \end{align}
Using Taylor approximation for $e^{-xy}$ respect $y$ at $y=0$ we get $e^{-xy}=1-xy+y\phi_x(y)$ where $\lim_{y\to 0}\phi_x(y)=0$, so
$$ u(x,y)=-\frac{f(y)}{y}e^{-xy}+g(y)=\left(g(y)-\frac{f(y)}{y}\right) + (xf(y) -\phi_x(y)f(y)) $$
Thus,
$$ A=u(0,0)=\lim_{y\to 0} u(0,y)=\lim_{y\to 0} \left(\left(g(y)-\frac{f(y)}{y}\right) -\phi_0(y)f(y)\right)= \lim_{y\to 0} \frac{yg(y)-f(y)}{y} $$
So if $h(y):=yg(y)-f(y) \in C^2(\mathbb{R}\setminus \{0\})$ we have $h'(0)=\lim_{y\to 0} \frac{h(y)}{y} =A$ and $\lim_{y\to 0}h(y)=0$ so taking $h(0):=0$ we have $h \in C(\mathbb{R})$. And for $y\neq 0$ we now have
$$ u(x,y)=\frac{1-e^{-xy}}{y}f(y)+\frac{h(y)}{y} $$
Now I have to impose conditions on $f$, $h$, $A$ so it is equivalent to $u\in C^2$
We now calculate $u_x$, $u_y$, $u_{xx}$, $u_{xy}$, $u_{yy}$ for $y\neq 0$
\begin{align} u_x(x,y)=&e^{-xy}f(y)\\ u_{xx}(x,y)=&-ye^{-xy}f(y)\\ u_y(x,y)=&-\frac{e^{-xy}-1}{y}f'(y)+\frac{1}{y}\left(xe^{-xy}+\frac{e^{-xy}-1}{y}\right)f(y)+\frac{h'(y)}{y}-\frac{h(y)}{y^2}\\ u_{xy}(x,y)=&(f'(y)-xf(y))e^{-xy}\\ u_{yy}(x,y)=&-\frac{e^{-xy}-1}{y}f''(y)+\frac{2((xy+1)e^{-xy}-1)}{y^2}f'(y)+\\ &+\frac{2-(xy(xy+2)+2)e^{-xy}}{y^3}f(y)+\frac{h''(y)}{y}-\frac{2h'(y)}{y^2}+\frac{2h(y)}{y^3} \end{align}
Now, because $u \in C^2(\mathbb{R}^2)$ $$\exists\lim_{y\to 0}u_{xy}(0,y)=\lim_{y\to 0}f'(y)$$
Then, $\exists f'(0):=\lim_{y\to 0}f'(y)$ so $f \in C^1(\mathbb{R})$. On the other hand,
$$\exists\lim_{y\to 0} u_y(0,y)=\lim_{y\to 0} \frac{h'(y)-\frac{h(y)}{y}}{y}\implies$$
$$\lim_{y \to 0} \left(h'(y)-\frac{h(y)}{y}\right)=0 \implies \lim_{y \to 0} h'(y) = h'(0)=A$$
since $\lim_{y\to 0} \frac{h(y)}{y} =A$, so $h\in C^1(\mathbb{R})$.