$$\frac{dy}{dx} = -3y; \quad (2,6)$$
This is what I've done so far.
$$ \frac{dy}{dx} = -3y $$ $$ dy = -3y \, dx $$ $$ dx = \frac{dy}{-3y} $$ $$ dx = dy \cdot \frac{1}{3} y $$ $$ \int dx = \frac{1}{3} \int y \,dy $$ $$ x = \frac{1}{3} \cdot \frac{y^2}{2} + c $$ $$ x = \frac{y^2}{6} + c $$
So at this point I should just plug $(2,6)$ in right? or is the whole working all messed up?
Your work should look like:$$\frac{dy}{dx}=-3y$$ $$\frac{dy}{y}=-3\,dx$$ Integrating both sides yeilds: $$\log y =-3x+C$$ Exponentiate both sides: $$y=Ke^{-3x}$$ where $K=e^C$
Since you are given initial condition $x=2$ and $y=6$, you can plug in those values to get:$$6=Ke^{-6}$$ So $$K=6e^6$$ and the full equation would be $$y=6e^{-3x+6}$$