Using the power series expansion is not so complicated if you notice that differentiating 32 times will eliminate any power of $x$ below 32, and setting $x=0$ will eliminate any term which has a power of $x$ greater than 32. There are high powers of $x$ involved so we do not need to look at many terms, and we can cut off the expansion once we get to $x^{32}$. It seems to me that just writing it out explicitly, leaving out irrelevant terms as you go, is as easy as any other method.
Do two stages using $R$ for an irrelevant remainder:
$\sin(x^{10}+x^{11}) = x^{10}+x^{11}-\frac {(x^{10}+x^{11})^3}{3!}+R = x^{10}+x^{11}-\frac {x^{30}} 6-\frac{x^{31}} 2-\frac{x^{32}} 2+R$
Now note that when this is exponentiated, the terms in $x^{30}$ and higher will become irrelevant when the whole expression is squared or cubed, so (ignoring irrelevant terms):
$e^{x^{10}+x^{11}-\frac {x^{30}} 6-\frac{x^{31}} 2-\frac{x^{32}} 2} = 1+\left(x^{10}+x^{11}-\frac {x^{30}} 6-\frac{x^{31}} 2-\frac{x^{32}} 2\right)+\frac 1 2 (x^{10}+x^{11})^2+\frac 1 6(x^{10}+x^{11})^3+R$
The only ultimately relevant term, as noted above, is the one in $x^{32}$, and picking this out from the terms we have we see that: $f(x)=-\frac{x^{32}}2+\frac{x^{32}}2+R$
The coefficient we are looking for vanishes, and the answer to the question is therefore 0.