y
   ∫(g(p))-2 dp = g(y), 
    0
and:
     4
   ∫(g(p))-2 dp = (12)1/3.
    0

merry christmas everyone :-)

let h(x)= (1/g(x))^2 then from the first equation we have

Integral h(t) dt from 0 to x  = g(x)

from the fundamental theorem of calculus we have that

d/dx Integral h(t) dt from 0 to x = h(x)

thus h(x)=g'(x)

thus we end up with the differential equation, whose solution is g(x)

dy/dx = (1/y)^2

y^2 dy = dx

integrating we get

y^3/3=x+c

thus y=(3x+c)^(1/3)

from the second equation given we get that

g(4)=12^(1/3)

thus

(12+c)^(1/3)=12^(1/3)

12+c=12

c=0

thus

g(x)=(3x)^(1/3)

QED