Find the derivative of the following

$\begin{aligned} x^ky^h & = (x+y)^{k+h} & \small \blue{\text{Differentiate both sides w.r.t. }x} \\ kx^{k-1}y^h + hx^ky^{h-1}\frac {dy}{dx} & = (k+h)\blue{(x+y)^{k+h-1}} \left(1+\frac {dy}{dx}\right) & \small \blue{\text{Note that }x^ky^h = (x+y)^{k+h}} \\ \frac kx x^ky^h + \frac hy x^ky^h\frac {dy}{dx} & = \frac {k+h}{\blue{x+y}}\blue{x^ky^h} \left(1+\frac {dy}{dx}\right) & \small \blue{\text{Divide both sides by }x^ky^h} \\ \frac kx + \frac hy \frac {dy}{dx} & = \frac {k+h}{x+y} \left(1+\frac {dy}{dx}\right) & \small \blue{\text{Rearrange}} \\ \left(\frac hy - \frac {k+h}{x+y} \right) \frac {dy}{dx} & = \frac {k+h}{x+y} - \frac kx \\ \left(\frac {hx-ky}{y(x+y)} \right) \frac {dy}{dx} & = \frac {hx-ky}{x(x+y)} \\ \implies \frac{dy}{dx} & = \boxed {\frac yx} \end{aligned}$

Let $y=vx$ . Substituting in the given equation, we have $x^k(vx)^h=(x+vx)^{k+h}$ , or $v^h=(v+1)^{k+h}$ . So $v$ is constant, independent of h and k . So $\dfrac{dy}{dx}=v=\dfrac{y}{x}$ .