When $m=1$, this problem becomes # 4294.

When $m\ne 1$, this expression can be tackled by the recursion method. Let the left side of this to-be-proved identity be $a_n$. It can be shown that $$a_n=\frac{n}{m+n}\cdot a_{n-1}$$

because $$\begin{align*} a_n =\ &\sum_{k=0}^{n}(-1)^k\frac{m}{m+k}\binom{n}{k} \\=\ & 1 + \sum_{k=1}^{n-1}(-1)^k\frac{m}{m+k}\left(\binom{n-1}{k} +\binom{n-1}{k-1}\right) + (-1)^n\frac{m}{n+m} \\=\ &\left(1 + \sum_{k=1}^{n-1}(-1)^k\frac{m}{m+k}\binom{n-1}{k}\right) \\ &\ + \left(\sum_{k=1}^{n-1}(-1)^k\frac{m}{m+k}\binom{n-1}{k-1} + (-1)^n\frac{m}{n+m}\right) \\=\ & a_{n-1} + \sum_{k=1}^{n}(-1)^k\frac{m}{m+k}\binom{n-1}{k-1} \\=\ & a_{n-1} +  \sum_{k=1}^{n}(-1)^k\frac{m}{m+k}\left(\frac{k}{n}\binom{n}{k}\right)\\=\ & a_{n-1} +  \sum_{k=0}^{n}(-1)^k\frac{m}{m+k}\left(\frac{k}{n}\binom{n}{k}\right)\\=\ & a_{n-1} + \frac{m}{n}\sum_{k=0}^{n}(-1)^k\frac{k}{m+k}\binom{n}{k}\\=\ & a_{n-1} + \frac{m}{n}\left(\sum_{k=0}^{n}(-1)^k\binom{n}{k}-\sum_{k=0}^{n}(-1)^k\frac{m}{m+k}\binom{n}{k}\right) \\=\ & a_{n-1} - \frac{m}{n}\left(0--\sum_{k=0}^{n}(-1)^k\frac{m}{m+k}\binom{n}{k}\right) \\=\ & a_{n-1}-\frac{m}{n}a_n\end{align*}$$

$$\therefore\quad a_n = a_{n-1} - \frac{m}{n}a_n \implies a_n = \frac{n}{m+n}a_{n-1}$$

It follows that $$\begin{align*} a_n =\ &\frac{n}{m+n}a_{n-1} \\ =\ & \frac{n}{m+n}\cdot\frac{n-1}{m+n-1} a_{n-2} \\ =\ & \cdots \\ =\ & \frac{n}{m+n}\cdot\frac{n-1}{m+n-1}\cdots\frac{1}{m+1}a_0 \\ =\ & \frac{m!n!}{(m+n)!}\cdot 1 \\=\ &\frac{1}{\binom{m+n}{n}}\end{align*}$$