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$\dfrac{dy}{dx}=e^y(e^x+x^2)\ \Rightarrow\ e^{-y}dy=(e^x+x^2)dx$
$\int e^{-y}dy=\int (e^x+x^2)dx$ $\Rightarrow\ -e^{-y}=e^x+\dfrac{x^3}{3}+C$
Hence, $e^{-y}+e^x+\dfrac{x^3}{3}=C$ (or $y=-\ln\!\big(C-e^x-\dfrac{x^3}{3}\big)$).
Answer: $\boxed{e^{-y}+e^x+\dfrac{x^3}{3}=C}$ ✅
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