Suppose we have a homogeneous demand function of degree 0: $x(\alpha p,\alpha w)=x(p,w)$, where $p=(p_1,p_2,\ldots,p_L)$ is a vector and $w$ is a scalar.
Differentiate $x(\alpha p,\alpha w)=x(p,w)$ with respect to $\alpha$.
And here I don't know how to differentiate. If we write $x(\alpha p,\alpha w)=x(p,w)$ as $x(u,v)=x(p,w)$ then using the chain rule we have that $$\sum_{i=1}^L \frac{\partial x(u,v)}{\partial u_i}\frac{\partial u_i}{\partial \alpha}+\frac{\partial x(u,v)}{\partial v}\frac{\partial v}{\partial \alpha}=0$$ $$\sum_{i=1}^L \frac{\partial x(u,v)}{\partial u_i}p_i+\frac{\partial x(u,v)}{\partial v}w=0$$
$$\sum_{i=1}^L \frac{\partial x(\alpha p,\alpha w)}{\partial u_i}p_i+\frac{\partial x(\alpha p,\alpha w)}{\partial v}w=0$$
But the answer is $$\sum_{i=1}^L \frac{\partial x(\alpha p,\alpha w)}{\partial p_i}p_i+\frac{\partial x(\alpha p,\alpha w)}{\partial w}w=0$$
The question is how $\partial u_i$ is substituted with $\partial p_i$ and $\partial v$ is substituted with $\partial w$
I guess for each $i\in\{1,\dots, L\}$ we have $$\frac{\partial x(\alpha p,\alpha w)}{\partial p_i}=\frac{\partial x(\alpha p,\alpha w)}{\partial (\alpha p_i)}\cdot \frac{d(\alpha p_i)}{d p_i}=\frac{\partial x(\alpha p,\alpha w)}{\partial u_i}\cdot \alpha.$$ Similarly we have $$\frac{\partial x(\alpha p,\alpha w)}{\partial w}=\frac{\partial x(\alpha p,\alpha w)}{\partial (\alpha w)}\cdot \frac{d(\alpha w)}{d w}=\frac{\partial x(\alpha p,\alpha w)}{\partial v}\cdot \alpha.$$
Then your answer differs from the answer by the factor $\frac 1\alpha$.
For instance, let $L=1$,$x(p_1,w)=\frac {p_1}w$ and $w,\alpha\ne 0$. Then $u_1=\alpha p_1$ and $v=\alpha w$. So
$$\frac{\partial x(\alpha p,\alpha w)}{\partial p_1}=\frac{\partial\frac{p_1}{w}}{\partial p_1}=\frac 1w\mbox{ and }\frac{\partial x(u,v)}{\partial u_1}=\frac{\partial\frac{u_1}{v}}{\partial u_1}=\frac 1v=\frac 1{\alpha w}.$$
Similarly we have
$$\frac{\partial x(\alpha p,\alpha w)}{\partial w}=\frac{\partial\frac{p_1}{w}}{\partial w}=-\frac {p_1}{w^2}\mbox{ and }\frac{\partial x(u,v)}{\partial v}=\frac{\partial\frac{u_1}{v}}{\partial v}=-\frac {u_1}{v^2}=-\frac {\alpha p_1}{(\alpha w)^2}.$$