For our purposes, we may assume that $f$ is continuous.
Step 1: $$ \phi '(x) = \bigg(\frac{d}{{\,dx}}\frac{1}{x}\bigg)\int_0^x {F(t)\,dt} + \frac{1}{x}\frac{d}{{\,dx}}\int_0^x {F(t)\,dt}. $$
Step 2: $$ \phi '(x) = - \frac{1}{{x^2 }}\int_0^x {F(t)\,dt} + \frac{1}{x}F(x). $$
Step 3: $$ x^2 \phi '(x) = - \int_0^x {F(t)dt} + xF(x) = xF(x) - \int_0^x {F(t)dt}. $$
Step 4 -- integration by parts: $$ \int_0^x {tf(t) \,dt} = tF(t) \big|_0^x - \int_0^x {\bigg(\frac{d}{{\,dt}}t \bigg)F(t)\,dt} = xF(x) - 0F(0) - \int_0^x {1F(t)\,dt} = xF(x) - \int_0^x {F(t)\,dt}. $$
Step 5: It follows that $$ x^2 \phi '(x) = \int_0^x {tf(t) \,dt}. $$
Step 6: It follows that $$ \phi '(x) = \frac{1}{{x^2 }}\int_0^x {tf(t) \,dt}. $$
EDIT: For the integration by parts (beginning of Step 4), note that $F$ is an antiderivative of $f$, since, by the Fundamental theorem of calculus, $$ F'(x) = \frac{d}{{dx}}F(x) = \frac{d}{{dx}}\int_0^x {f(t)dt} = f(x). $$ (Here we used the assumption that $f$ is continuous.)