Y^2 = xtan(2-y) find dy/dx

please help me figure it out

y^2 = x tan(2 - y)

2y dy/dx = x sec^2(2-y)(-dy/dx) + tan(2 - y)

dy/dx[ 2y + x sec^2(2 - y)] = tan(2 - y)

dy/dx = tan(2 - y) / [2y + x sec^2(2 - y) ]