Exists $t^*in mathbb{R}$ such that $y(t^*)=-1$?.

First, as $y^2-13y+77$ is of class $C^1$ on $y$, we have existence and uniqueness of the solution.

Note that $y'(t) = y^2-13y+77 = (y-frac{13}{2})^2+frac{477}{4}ge frac{477}{4}$, for all $tinmathbb{R}$. If there no exists a $t^*$ with $y(t^*)=-1$ then, by continuity of the solution, $y(t)>-1$, for all $tinmathbb{R}$.

By the Mean Value Theorem, for every $t>0$, there exists a $cin(-1,frac{13}{2})$, such that $$ y'(c) = frac{y(t)-frac{13}{2}}{t-0}$$

As $y(t)in (-1,frac{13}{2})$, you can choice $t_0$ large enough such that $frac{y(t)-frac{13}{2}}{t-0}<1$. So there exists a $c$ with $y'(c)<1$, but it contradicts $y'(t)gefrac{477}{4}$.

So there exists a $t^*$ such that $y(t^*)=-1$.

In this case the situation is more involved (as compared with the other post). The r.h.s. is always strictly positive, hence the solution is strictly increasing.

By separation of variables, its implicit form is $$ F(y) = t, qquad F(y) := int_{13/2}^y frac{1}{s^2-13s+77}, ds. $$ ($F$ can be explicitly computed, but there is no need to do the computation.)

By studying the convergence of the integral, there exist (and are finite) the limits $$ T_1 := lim_{yto + infty} F(y), qquad T_0 := lim_{yto - infty} F(y). $$ Hence, your solution is defined for $tin (T_0, T_1)$; moreover, it can be proved that $$ lim_{tto T_0+} y(t) = -infty, qquad lim_{tto T_1-} y(t) = +infty, $$ so that the image of $y$ is $mathbb{R}$.