If $f$ is homogeneous of degree $n$, set $p = xt$ and $q = yt$ and define $h(x,y,t) = f(p,q) = t^nf(x,y)$ A function $f$:$\mathbb R$ $\rightarrow$ $\mathbb R$  is called homogeneous of degree n if it satisfies $f(tx,ty) = t^nf(x,y)$
If $f$ is homogeneous of degree $n$, set $p = xt$ and $q = yt$ and define
$h(x,y,t) = f(p,q) = t^nf(x,y)$
Apply the chain rule to $h(x,y,t)$ to show that

$x\frac {\partial {f}}{dx} + y\frac {\partial {f}}{dy} = nf(x,y)$

I don't know how to start, usually with homogeneous equation I substitute x and y with tx and ty and gives me this  $f(tx,ty) = t^nf(x,y)$ form with degree. I'm not quite sure how to tackle this question.

I think I got it, Let me know if I'm right
$h(x,y,t) = f(p,q) = t^nf(x,y)$
$f(xt,yt) = t^nf(x,y)$ we were given If $f$ is homogeneous of degree $n$, set $p = xt$ and $q = yt$
Now take the partial derivative on L.H.S and full derivative on R.H.S
$xt\frac {\partial {f}}{dxt} + yt\frac {\partial {f}}{dyt} = nt^{n-1}f(x,y)$
Now set $t = 1$
$x\frac {\partial {f}}{dx} + y\frac {\partial {f}}{dy} = nf(x,y)$
