First bring the equation in the form

$$A(t,y) y'(t) + B(t,y) =0$$

just by putting the parenthesis in a different manner you instantly get:

$$A(t,y)= q(t,y) (t^2+1)\sin(y), \qquad B(t,y) = q(t,y) t$$

for exactness a necessary condition is that $\partial_t A = \partial_y B$. We evaluate

$$\partial_t A = \partial_t q(t,y) (t^2+1) \sin(y) + 2 q(t,y) t \sin(y),\qquad \partial_y B = \partial_y q(t,y) t$$

you want these two to be equal. That is probably annoying to solve in general, but we are only interested in finding one solution. So first we just try find a $q$ that solves $\partial_t A = 0$ and $\partial_y B=0$, if we find such a solution then clearly $\partial_t A = \partial_y B$. The first equation becomes:

$$\partial_t A = 0 = \sin(y) ( \partial_t q (t^2+1) +2t q)$$

divide by $\sin(y)$ and $q$ and $(t^2+1)$ to get:

$$0=\frac{\partial_t q}{q}+\frac{2t}{t^2+1}$$

by noting $\frac{\partial_t q}q = \partial_t \ln(q)$ you get want to find a $q$ so that

$$\partial_t \ln(q) = -\frac{2t}{1+t^2}$$

one solution is found by integrating the right-hand side:

$$\ln(q) = -\frac12 \ln(1+t^2)$$

so

$$q= \frac1{\sqrt{1+t^2}}$$

Now check by hand again just to be safe that $\partial_t A =0$, this is true. We also see that $\partial_y B = 0$. So this solves the necessary condition for exactness that I gave above.

In many lectures this necessary condition is actually the definition, so you would be done here if that is the case. In other lectures the ODE is exact if there is an $F(t,y)$ so that $\partial_y F = A$ and $\partial_t F = B$. It is easy to check that for example $F = -\cos(y)\sqrt{1+t^2}$ is a solution for $q=\frac1{\sqrt{1+t^2}}$ - so the calculation above that found the $q$ is really the only calculation to do here.

Low bounty!