Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,1} \right)} \frac{{y\sin x}}{{x\left( {y + 1} \right)}} \cr
& {\text{Use the product property of limits}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,1} \right)} \frac{{y\sin x}}{{x\left( {y + 1} \right)}} = \left[ {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,1} \right)} \frac{{\sin x}}{x}} \right]\left[ {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,1} \right)} \frac{y}{{y + 1}}} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}} \right]\left[ {\mathop {\lim }\limits_{y \to 1} \frac{y}{{y + 1}}} \right] \cr
& {\text{Evaluating the limits}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,1} \right)} \frac{{y\sin x}}{{x\left( {y + 1} \right)}} = \left[ 1 \right]\left[ {\frac{1}{{1 + 1}}} \right] \cr
& {\text{Simplifying}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,1} \right)} \frac{{y\sin x}}{{x\left( {y + 1} \right)}} = \frac{1}{2} \cr} $$
