Let $f$ is a differentiable function such that $f(x)=x^2+\int_0^x e^{-t} f(x-t) d t$, then:

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We have $f(x)=x^{2}+\int_{0}^{x}e^{-t}f(x-t)dt$ $\Rightarrow f(x)=x^{2}+\int_{0}^{x}e^{-(x-t)}f(x-(x-t))dt$ $[u~sin~g\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx]$ $\Rightarrow f(x)=x^{2}+e^{-x}\int_{0}^{x}e^{t}f(t)dt...(i)$ On differentiating both the sides, get $f^{\prime}(x)=2x+e^{-x}[e^{x}f(x)]-e^{-x}\int_{0}^{x}e^{t}f(t)dt$ [using Leibnitz rule] ...(ii) $\Rightarrow f(x)+f^{\prime}(x)=x^{2}+2x+f(x)\Rightarrow f^{\prime}(x)=x^{2}+2x$ ...(iii) On integrating both the sides of equation (iii), we get $f(x)=\frac{x^{3}}{3}+x^{2}+C$ But $f(0)=0\Rightarrow C=0$ Hence, $f(x)=\frac{x^{3}}{3}+x^{2}$