$y=f(x)$ is

$\begin{aligned}f(x)=& x^{2}+\int_{0}^{x} e^{-t} f(x-t) d t \\=& x^{2}+\int_{0}^{x} e^{-(x-t)} f(x-(x-t)) d t \\&=x^{2}+e^{-x} \int_{0}^{x} e^{t} f(t) d t\end{aligned}$

Differentiating w.r.t. $x,$ we get

$\begin{aligned}\Rightarrow f^{\prime}(x)=& 2 x-e^{-x} \int_{0}^{x} e^{t} f(t) d t+e^{-x} \cdot e^{x} f(x) \\&=2 x-e^{-x} \int_{0}^{x} e^{t} f(t) d t+f(x)\end{aligned}$

$\Rightarrow f^{\prime}(x)=2 x+x^{2} \\$

$\Rightarrow f(x)=\dfrac{x^{3}}{3}+x^{2}+c\\$

Also $f(0)=0\\$ [using equation from equation]

$\Rightarrow f(x)=\dfrac{x^{3}}{3}+x^{2} \\$

$\Rightarrow f^{\prime}(x)=x^{2}+2 x\\$

$\Rightarrow f^{\prime}(x)=0$ has real roots, hence $f^{\prime}(x)$ is non-monotonic

Hence, $f(x)$ is many-one, but range is $R$, hence surjective.