The answer is given below :

Given,

x^m y^n = (x + y)^m

Now, taking log to both sides, we get

log (x^m y^n) = log (x + y)^(m + n)

⇒ log (x^m) + log (y^n) = log (x + y)^(m + n),
since log (ab) = log a + log B

⇒ m LOGX + n log y = (m + n) log (x + y),
since log (a^b) = b log a

Now, differentiating both sides with respect to x, we get

d/dx (m logx) + d/dx (n logy)
= d/dx [(m + n) log (x + y)]

⇒ m d/dx (logx) + n d/dx (logy)
= (m + n) d/dx [log (x + y)],
since d/dx [k f(x)] = k d/dx [f(x)], where k is a constant

⇒ m/x + n/y dy/dx = (m + n)/(x + y) d/dx (x + y),
since d/dx (log x) = 1/x

⇒ m/x + n/y dy/dx = (m + n)/(x + y) (1 + dy/dx)

⇒ m/x + n/y dy/dx = (m + n)/(x + y)
+ (m + n)/(x + y) dy/dx

⇒ n/y dy/dx - (m + n)/(x + y) dy/dx
= (m + n)/(x + y) - m/x

⇒ [n/y - (m + n)/(x + y)] dy/dx
= [(m + n)/(x + y) - m/x]

⇒ [{n(x + y) - (m + n)y}/{y(x + y)}] dy/dx
= {x(m + n) - m(x + y)}/{x(x + y)}

⇒ (nx + ny - my - ny)/{y(x + y)} dy/dx
= (mx + nx - mx - my)/{x(x + y)}

⇒ (nx - my)/(x + y) 1/y dy/dx
= (nx - my)/(x + y) 1/x

⇒ 1/y dy/dx = 1/x

⇒ dy/dx = y/x [Proved]

THANK you for your QUESTION.