simply let y be the amount of salt in the tank at time t

dy/dt = salt in -salt out

the volume of fluid in the tank is v(t) = 60 -t gals

dy/dt =(1lb/gal)(2gal/min) - (ylbs/60 -t gals)(3gals/min)

dy/dt = 1 - 3y/(60 -t) y(0) = 0

This is not separable so rewrite

dy/dt + 3y /(60-t) = 1 y(0) = 0

Now use an integrating factor to solve