Solving $-u''(x) = \delta(x)$

This is a good example of a question to which one can answer at some very different levels of mathematical sophistication... Since you say nothing about this, let me try an elementary approach.

What you call the Dirac delta function (which is not a function, at least not in the sense of a function from $\mathbb R$ to $\mathbb R$) is a strange object but something about it is clear:

One asks that $\displaystyle\int_y^z\delta(x)\mathrm dx=0$ if $y\leqslant z<0$ or if $0<y\leqslant z$ and that $\displaystyle\int_y^z\delta(x)\mathrm dx=1$ is $y<0<z$.

We will not use anything else about the Dirac $\delta$.

If one also asks that $\displaystyle\int_y^zu''(x)\mathrm dx=u'(z)-u'(y)$ for every $y\leqslant z$, one can integrate once your equation $\color{red}{-u''=\delta}$, getting that there exists $a$ such that $$ u'(x)=a-[x\geqslant0], $$ where we used Iverson bracket notation. Now let us integrate this once again.

Using the facts that $\displaystyle\int_y^zu'(x)\mathrm dx$ should be $u(z)-u(y)$ for every $y\leqslant z$, and the value of $\displaystyle\int_y^z[x\geqslant0]\mathrm dx$, one gets that for every fixed negative number $x_0$, $$ u(x)=u(x_0)+a\cdot (x-x_0)-x\cdot[x\geqslant0]. $$ This means that $b=u(x_0)-a\cdot x_0$ does not depend on $x_0<0$, hence finally, for every $x$ in $\mathbb R$, $$ \color{red}{u(x)=a\cdot x+b-x\cdot[x\geqslant0]}. $$ (And, in the present case, the condition that $u(-2)=u(3)=0$ imposes that $a=3/5$ and $b=6/5$.)

This is the general solution of the equation $-u''=\delta$. Note that every solution $u$ is $C^\infty$ on $\mathbb R\setminus\{0\}$ but only $C^0$ at $0$ hence $u'$ and $u''$ do not exist in the rigorous sense usually meant in mathematics. Note finally that $u$ is also $$ u(x)=a\cdot x+b-x\cdot[x\gt0]. $$

Both describe the same type of function. The ramp function is nothing but \begin{align} R(x) = \begin{cases}0 & x \leq 0 \\ x & x \geq 0\end{cases} \end{align} If you use the general solution and plug in the same boundary conditions, \begin{align}u(-2) &= -R(-2) + C -2D = C - 2D = 0 \\ u(3) &= -R(3) + C + 3D = -3 + C +3D = 0\end{align} with the solution $C=6/5$, $D=3/5$, and then split it at $x=0$ to get rid of the ramp function and you obtain; \begin{align}u(x) = \begin{cases} \frac65 + \frac35 x & x \leq 0\\ -x + \frac65 + \frac35 x = \frac65 - \frac25 x &x \geq 0\end{cases}\end{align}

which is exactly the same expression you got by splitting the function earlier (pratically, there is no difference, but its shorter when written with the ramp function).