find a function f such that f'(x)=x^{3} and the line x+y=0 is tangent to the graph of f
$\displaystyle f'(x) = x^3 \implies f(x) = \frac{x^4}{4} + C$
$\displaystyle x+y = 0 \implies y = -x$ ... slope of the tangent line is $\displaystyle m = -1$
at the point of tangency $\displaystyle (x_0,y_0)$ ...
$\displaystyle f'(x_0) = x_0^3 = -1 \implies x_0 = -1$
finally ...
$\displaystyle y_0 = -x_0 = -(-1) = \frac{(-1)^4}{4} + C$
... solve for C to finish it.