If $f(x) = x^{(x+1)}(x+2)^{(x+3)}$, then $f(0)+f(-1)+f(-2)+f(-3) =$
$\text{(A)}\ -\frac{8}{9}\qquad\text{(B)}\ 0\qquad\text{(C)}\ \frac{8}{9}\qquad\text{(D)}\ 1\qquad\text{(E)}\ \frac{10}{9}$
Plugging in $x=0$ into the function will give $0^1\cdot 2^3$.  Since $0^1 = 0$, this gives $0$.
Plugging in $x=-1$ into the function will give $(-1)^0 \cdot 1^2$.  Since $(-1)^0 = 1$ and $1^2 = 1$, this gives $1$.
Plugging in $x=-2$ will give a $0^1$ factor as the second term, giving an answer of $0$.
Plugging in $x=-3$ will give $(-3)^{-2}\cdot (-1)^0$.  The last term is $1$, while the first term is $\frac{1}{(-3)^2} = \frac{1}{9}$
Adding up all four values, the answer is $1 + \frac{1}{9} = \frac{10}{9}$, and the right answer is $\boxed{E}$.