$$P(s)=\int_{0}^{1}\int_{0}^{1}xy \, \delta{(s-(x+y))}\, dxdy$$
$P(s)$ is the probability density of random variable $s=x+y$, who is a function of the two original random variables $x,y$. The peak occurs inside the limit of integration.
$P(x,y)=xy$ is the probability density of random variable $x,y$.
I tried to apply the property of Dirac delta function $$\int_{a}^{b}f(x)\, \delta{(x-x_{0})}\, dx=f(x_{0}).$$
However, I am having trouble identify $x_{0}$.
Edit: screenshot below
This is the example the textbook use. For the exercise, $P(x,y)=xy$ instead of $1/36$, and I am kinda lost here.
Here is a pedestrian approach: $$\begin{align}P(s)~=~&\int_{[0,1]}\!dy~y \int_{[0,1]}\!dx~x~\delta{(s-(x+y))} \cr ~=~&\int_{[0,1]}\!dy~y \int_{\mathbb{R}}dx ~x~1_{[0,1]}(x) ~ \delta{(s-y-x)}\cr ~=~&\int_{[0,1]}\!dy~y(s-y)~1_{[0,1]}(s-y) \cr ~=~&\int_{\mathbb{R}}dy ~y(s-y)~1_{[0,1]}(y)~1_{[s-1,s]}(y) \cr ~=~&\int_{\mathbb{R}}dy ~y(s-y)~1_{[\max(0,s-1),\min(s,1)]}(y) \cr ~=~&\int_{\mathbb{R}}dy ~y(s-y)~\left(1_{[0,1]}(s) 1_{[0,s]}(y) + 1_{[1,2]}(s) 1_{[s-1,1]}(y) \right) \cr ~=~&1_{[0,1]}(s) \left[ ~y^2\left(\frac{s}{2}-\frac{y}{3}\right)\right]^s_0 + 1_{[1,2]}(s) \left[ ~y^2\left(\frac{s}{2}-\frac{y}{3}\right)\right]^1_{s-1} \cr ~=~&1_{[0,1]}(s)~\frac{s^3}{6} + 1_{[1,2]}(s) \left( -\frac{s^3}{6}+s-\frac{2}{3}\right), \end{align}$$ where we have used indicator/characteristic functions.
From the 3rd line one can see that the map $s\mapsto P(s)$ is continuous. The last expression is continuous if the value the of indicator function is $1/2$ at the interval endpoints.