Let $f$ be a homogenous function of degree $k$ differentiable away from the origin. Show that for any unit vector $u$ the directional derivative is homogeneous of degree $k-1$.
(Incorrect) Proof Let $u$ be given. Since $f$ is differentiable, $\partial_uf(x)$ exists.
$\begin{equation} \begin{split} \partial_uf(\lambda x) &= \lim_{h \to 0} \cfrac{f(\lambda x + uh) - f(\lambda x)}{h}\\ &= \lim_{h \to 0} \cfrac{f(\lambda (x + \cfrac{h}{\lambda}u) - f(\lambda x)}{h}\\ &= \lim_{h \to 0} \cfrac{\lambda^k\big[f(x + \cfrac{h}{\lambda}u) - f(x)\big ]}{h} \\ &= \lambda^k \partial_uf(x) \end{split} \end{equation}$
Where do I go wrong?
You went wrong at the last step. To recognize the limit as the directional derivative, you need $h/\lambda$ in the denominator, and to get that you "borrow" one power of $\lambda$ from the $\lambda^k$. (Once you have $h/\lambda$ in the denominator, you do the standard thing of substituting $k=h/\lambda$ and noting that $h\to 0$ if and only if $k\to 0$.)