Answer
$$y=\frac{1}{3 \pi }\sin (3\pi t)-\frac{1}{4\pi }\cos (4\pi t) - \frac{\pi}{8}$$
Work Step by Step
Given$$ \frac{d y}{d t}=\cos (3\pi t)+\sin (4\pi t) , \quad y(1/3)= 0 $$
Since :\begin{align*}
y&= \int (\cos (3\pi t)+\sin (4\pi t) ) dt\\
&=\frac{1}{3 \pi }\sin (3\pi t)-\frac{1}{4\pi }\cos (4\pi t) +c
\end{align*}
at $t=\frac{1}{3},\ \ \ y= 0 $, then $c= -\pi/8$, hence:
$$y=\frac{1}{3 \pi }\sin (3\pi t)-\frac{1}{4\pi }\cos (4\pi t) - \frac{\pi}{8}$$