We don't need L'Hospital's rule since we can write $f_n(x)=n\int_0^{\frac{x^2}n}e^tdt=\int_0^{x^2}e^{\frac yn}dy,$ making the substitution $y=nt$. Hence for a fixed $x$ the sequence $\{f_n(x)\}$ is decreasing. Furthermore, the convergence is uniform on all compact set $\left[-A,A\right]$ since $\sup_{x\in \left[-A,A\right]}|f_n(x)-x^2|=\int_0^{x^2}\underbrace{(e^{\frac yn}-1)}_{\geq 0}dy\leq \int_0^{A^2}\left(\int_0^{\frac yn}e^tdt\right)dy,$ and making the substitution $u=nt$, we get $\sup_{x\in \left[-A,A\right]}|f_n(x)-x^2|\leq \int_0^{A^2}\left(\int_0^ye^{\frac un}\frac{du}n\right)dy\leq\frac 1n\int_0^{A^2}ye^{\frac yn}dy\leq \frac Ane^{\frac An}.$ Since $f_n(\sqrt n)=n(e-1)$, we can't hope more.