Rewrite:
(4+2y)⅔=(2.52/1.875)y5/3, cube both sides:
(4+2y)2=(2.52/1.875)3y5; let a=(2.52/1.875)3=0.60693 approx.
4+4y+y2=ay5; ay5-y2-4y-4=0.
Let f(y)=ay5-y2-4y-4, f'(y)=5ay4-2y-4.
Using Newton's Method, yn+1=yn-f(yn)/f'(yn), let y0=2.
y1=1.9156..., y2=1.9058..., y3=1.9056..., y4=1.905686128..., y5=1.905686128457...
So the approximate solution is y=1.905686128457.