I want to solve the following ode of distribution, $$\partial_y^2 f(x,y)=\delta_x,$$ where $\delta_x$ is the Dirac delta function defined by $\left<\delta_x,g(y)\right>=g(x).$
I did some attempts, $$\left<\partial_y^2 f(x,y),\phi(y)\right>=\left<f(x,y),\partial_y^2\phi(y)\right>=\phi(x).$$
But I can't get the explicit formula of $f(x,y).$
You can either use the the well known and easy to proof formulas: $$ \partial_yH(y-x)=\delta_x\\ \partial_y(|y-x|)=sgn(y-x) $$ where the equality is to be understood in a distributional sense. With these formulas you can straightforward integrate the ODE w.r.t. $y$ twice.
Alternatively, you can (in a first step) restrict yourself to test functions whose (compact) support lies entirely above or below $x$. The ODE for such test functions obviously reads $$ \partial_y^2f(x,y)=0 $$ Thus, we get a potential full solution by $$ f=\begin{cases} yg_-(x)+h_-(x)&y<x\\ yg_+(x)+h_+(x)&y<x \end{cases} $$ Now you simply need to fit $g_\pm, h_\pm$ such that $f$ is continous and its second distributional derivate yields the $\delta_x$.
Either way you end up with something like this $$ f=\frac{1}{2}|y-x| + y g(x) + h(x) $$