Which is the Taylor series of the second order, centered at the origin, of the function:
$$ f(x,y)=(1+x^3)^y $$
You might write your function as $ f(x,y) = \exp(y \log(1+x^3))$. Now $\log(1+x^3) = O(x^3)$, so $$f(x,y) = \exp(O(x^3 y)) = 1 + O(x^3 y)$$ Thus the Taylor expansion to second order is just $1$.
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You might write your function as $ f(x,y) = \exp(y \log(1+x^3))$. Now $\log(1+x^3) = O(x^3)$, so $$f(x,y) = \exp(O(x^3 y)) = 1 + O(x^3 y)$$ Thus the Taylor expansion to second order is just $1$.