The equation is $$v_f^2=v_i^2+2a\Delta y$$
Starting with
$$\frac{dv}{dt}=a$$
I was told the first step is to multiply both sides by dy. However I'm not sure where this dy is coming from when we are given dV/dt = a.
Any help starting this would be greatly appreciated.
Note that $\frac{dv}{dt}=a \implies 2v \frac{dv}{dt} = 2av$. But $v = \frac{dy}{dt}$. Therefore, $$\int_{t_0}^{t_f} 2v \frac{dv}{dt} dt = 2a \int_{t_0}^{t_f} \frac{dy}{dt} dt.$$ Consequently, $$v(t)^2 \large|_{t_0}^{t_f} = v_f^2-v_i^2 = 2a y(t) \large|_{t_0}^{t_f} = 2a \Delta y.$$