Derivative of product of three functions: product rule
To find the derivative of $(abc)'$ you use repeated application of the product rule: $$ (abc)' = (ab)'c+abc' = (ab'+a'b)c+abc' = a'bc+ab'c+abc'. $$ In your case $a(x) = x$, $b(x) = \mathrm e^x$ and $c(x) = \operatorname{csc}(x)$, so $$ a' = 1, b' = \mathrm e^x \text{ and }c' = -\cot x\csc x. $$
To make it more clear: in $x \mathrm e^x\csc x$ you have three function rather than two, but $x\mathrm e^x$ is also a product of two functions, so $$ (x\mathrm e^x\csc x)' = (x\mathrm e^x)'\csc x+x\mathrm e^x(\csc x)'. $$ We can calculate the latter term, but what about $(x\mathrm e^x)'$? You again apply the product rule: $$ (x\mathrm e^x)' = x'\mathrm e^x+x(\mathrm e^x)'. $$
As J.M. said, one easy way yo handle arbitrary products and quotients is to use the logarithmic derivative: $(\ln f(x))' =\dfrac{f'(x)}{f(x)} $.
Suppose $f(x)$ contains a mixture of products and quotients: $f(x) =\dfrac{a_1(x)a_2(x) ... a_m(x)}{b_1(x)b_2(x)...b_n(x)} =\dfrac{\prod_{i=1}^m a_i(x)}{\prod_{j=1}^n b_j(x)} $.
Taking the log, $\ln f(x) =\sum_{i=1}^m \ln a_i(x)-\sum_{j=1}^n \ln b_j(x) $.
Differentiating both sides,
$\begin{array}\\ \dfrac{f'(x)}{f(x)} &=(\ln f(x))'\\ &=(\sum_{i=1}^m \ln a_i(x)-\sum_{j=1}^n \ln b_j(x))'\\ &=\sum_{i=1}^m (\ln a_i(x))'-\sum_{j=1}^n (\ln b_j(x))'\\ &=\sum_{i=1}^m \dfrac{a_i'(x)}{a_i(x)}-\sum_{j=1}^n \dfrac{b_j'(x)}{b_j(x)}\\ \end{array} $.
Therefore
$\begin{array}\\ f'(x) &=f(x)\left(\sum_{i=1}^m \dfrac{a_i'(x)}{a_i(x)}-\sum_{j=1}^n \dfrac{b_j'(x)}{b_j(x)}\right)\\ &=\dfrac{\prod_{i=1}^m a_i(x)}{\prod_{j=1}^n b_j(x)}\left(\sum_{i=1}^m \dfrac{a_i'(x)}{a_i(x)}-\sum_{j=1}^n \dfrac{b_j'(x)}{b_j(x)}\right)\\ \end{array} $.
Here are the two most common cases:
$f(x) = a_1(x) a_2(x)$ - the product with $m=2$ and $n=0$.
$f'(x) =a_1(x) a_2(x)\left(\dfrac{a_1'(x)}{a_1(x)}+\dfrac{a_2'(x)}{a_2(x)}\right) =a_1'(x) a_2(x)+a_1(x) a_2'(x) $.
$f(x) = \dfrac{a(x)}{b(x)}$ - the quotient with $m=n=1$.
$f'(x) = \dfrac{a(x)}{b(x)}\left(\dfrac{a'(x)}{a(x)}-\dfrac{b'(x)}{b(x)}\right) = \dfrac{a'(x)}{b(x)}-\dfrac{a(x)b'(x)}{b^2(x)} = \dfrac{a'(x)b(x)-a(x)b'(x)}{b^2(x)} $.
Finally, your case - the product of three functions:
$\begin{array}\\ f(x) &=a(x)b(x)c(x)\left(\dfrac{a'(x)}{a(x)}+\dfrac{b'(x)}{b(x)}+\dfrac{c'(x)}{c(x)}\right)\\ &=a'(x)b(x)c(x)+a(x)b'(x)c(x)+a(x)b(x)c'(x)\\ \end{array} $