Answer
$f(x)=\frac{1}{4}x^4+\frac{3}{4}$
Work Step by Step
In slope-intercept form, the equation of the tangent line is $y=-x$. Hence, the slope is $-1$. Since $f'(x)=x^3,$ we can plug in that slope to find the x-coordinate of the point at which the line is tangent to. $$-1=x^3 \Rightarrow x=-1.$$ Plugging that into the tangent line equation, we find that $y=1$. Now we just need to find the antiderivative of $x^3$ such that it hits the point $(-1,1)$.
$$f(x)=\frac{1}{4}x^4+C$$
$$1=\frac{1}{4}\cdot{-1}^4+C \Rightarrow C=\frac{3}{4}$$