Given x^p * y^q =(x+y)^p+q

Take log on both side

log x^p + log y^q = log(x+y)^p+q

=> plog x + qlogy = (p+q)*log (x+y)

Now differentiate with respect to x

      p/x + (q/y)* dy/dx = {(p+q)/(x + y)}*(1 + dy/dx)

=> p/x + (q/y)* dy/dx = {(p+q)/(x + y) + {(p+q)/(x + y)}*(dy/dx)

=> (q/y)* dy/dx - {(p+q)/(x + y)}*(dy/dx) = (p+q)/(x + y) - p/x

=> {(q/y) - (p+q)/(x + y)}*(dy/dx) = (p+q)/(x + y) - p/x 

=> [(qx + qy - py - qy)/{y*(x + y)}]*(dy/dx) = (px + qx - px - py)/{x*(x+y)}

=> {(qx - py)/y}*(dy/dx) = (qx - py)/x

=> (dy/dx)/y = 1/x

=> dy/dx = y/x

Hense proved.