Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,0} \right)} \frac{{y\ln y}}{x} \cr
& {\text{Use the product property of limits}} \cr
& = \left( {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,0} \right)} \frac{1}{x}} \right)\left( {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,0} \right)} y\ln y} \right) \cr
& = \left( {\mathop {\lim }\limits_{x \to 1} \frac{1}{x}} \right)\left( {\mathop {\lim }\limits_{y \to 0} y\ln y} \right) \cr
& {\text{Write }}\mathop {\lim }\limits_{y \to 0} y\ln y{\text{ as }}\mathop {\lim }\limits_{y \to 0} \frac{{\ln y}}{{1/y}} \cr
& = \left( {\mathop {\lim }\limits_{x \to 1} \frac{1}{x}} \right)\left( {\mathop {\lim }\limits_{y \to 0} \frac{{\ln y}}{{1/y}}} \right) \cr
& {\text{Calculating }}\mathop {\lim }\limits_{y \to 0} \frac{{\ln y}}{{1/y}}{\text{ by using the L'Hopitals Rule}} \cr
& = \left( {\mathop {\lim }\limits_{x \to 1} \frac{1}{x}} \right)\mathop {\lim }\limits_{y \to 0} \left( {\frac{{\frac{1}{y}}}{{ - \frac{1}{{{y^2}}}}}} \right) \cr
& = \left( {\mathop {\lim }\limits_{x \to 1} \frac{1}{x}} \right)\mathop {\lim }\limits_{y \to 0} \left( { - y} \right) \cr
& {\text{Evaluating the limits}} \cr
& = \left( {\frac{1}{1}} \right)\left( { - 0} \right) \cr
& = 0 \cr} $$
