$\displaystyle E=\sum_{k=0}^{m}\binom{m}{k}\dfrac{k\lambda ^k\mu ^{m-k}}{(\lambda +\mu)^m}=\dfrac{1}{(\lambda +\mu)^m}\sum_{k=0}^{m}\binom{m}{k}k\lambda ^k\mu ^{m-k}$
So the problem is how to calculate this series.
$$\sum_{k=0}^{m}\binom{m}{k}k\lambda ^k\mu ^{m-k}$$
The whole problem is shown as below.
Two independent random variables X and Y are Poisson random variables with parameter $\lambda$ and $\mu$,$\quad$For any non-negative integer $k\leq m$,what is $\mathbb{E}[X|X+Y=m]$
$$ \lambda\sum_{k=0}^{m}\binom{m}{k}k\lambda ^{k-1}\mu ^{m-k} = \lambda\sum_{k=0}^{m}\binom{m}{k}\frac{\partial}{\partial \lambda}\lambda ^{k}\mu ^{m-k} = \lambda\frac{\partial}{\partial \lambda}\sum_{k=0}^{m}\binom{m}{k}\lambda ^{k}\mu ^{m-k} $$ we can use $$ (x + y)^m = \sum_{k=0}^{m}\binom{m}{k}x^{m-k}y^k $$ or $$ \lambda\frac{\partial}{\partial \lambda}\sum_{k=0}^{m}\binom{m}{k}\lambda ^{k}\mu ^{m-k} = \lambda\frac{\partial}{\partial \lambda}(\mu + \lambda)^m $$ this leads $$ \sum_{k=0}^{m}\binom{m}{k}k\lambda ^{k}\mu ^{m-k} = m\lambda(\mu + \lambda)^{m-1} $$
$$\sum_{k=1}^{m}\binom{m}{k}k\lambda^k\mu^{m-k}=\sum_{k=1}^{m}m\binom{m-1}{k-1}\lambda^k\mu^{m-k}=m\lambda\sum_{k=0}^{m-1}\binom{m-1}{k}\lambda^k\mu^{(m-1)-k}=m\lambda(\lambda+\mu)^{m-1}.$$
It's better to work with your original expression: $$ E=\sum_{k=0}^{m}\binom{m}{k}\dfrac{k\lambda ^k\mu ^{m-k}}{(\lambda +\mu)^m}=\sum_{k=0}^m\binom m k k \left(\frac\lambda{\lambda+\mu}\right)^k \left(\frac \mu{\lambda+\mu}\right)^{m-k}. $$ Recognize that $\frac\lambda{\lambda+\mu}$ and $\frac\mu{\lambda+\mu}$ sum to unity; thus the $E$ you are seeking is the expectation for a binomial random variable with $m$ trials and success probability $\frac\lambda{\lambda+\mu}$ and hence equals $\frac{m\lambda}{\lambda+\mu}$.
This is not a coincidence: you can show that when $X$ and $Y$ are independent Poisson variables with rate parameters $\lambda$ and $\mu$ respecitvely, the distribution of $X$ conditional on $X+Y=m$ is binomial with $m$ trials and success probability $\lambda/(\lambda + \mu)$
