Answer
$$y'\left( x \right) = - \frac{{xy}}{{{x^2} + 2{y^2}}}$$
Work Step by Step
$$\eqalign{
& y\sqrt {{x^2} + {y^2}} = 15 \cr
& {\text{use implicit differentiation differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {y\sqrt {{x^2} + {y^2}} } \right] = \frac{d}{{dx}}\left[ {15} \right] \cr
& {\text{use product rule in the left side}} \cr
& y\frac{d}{{dx}}\left[ {\sqrt {{x^2} + {y^2}} } \right] + \sqrt {{x^2} + {y^2}} \frac{d}{{dx}}\left[ y \right] = \frac{d}{{dx}}\left[ {15} \right] \cr
& {\text{solving derivatives}} \cr
& y\left( {\frac{{\frac{d}{{dx}}\left[ {{x^2} + {y^2}} \right]}}{{2\sqrt {{x^2} + {y^2}} }}} \right) + \sqrt {{x^2} + {y^2}} y'\left( x \right) = 0 \cr
& y\left( {\frac{{2x + 2yy'\left( x \right)}}{{2\sqrt {{x^2} + {y^2}} }}} \right) + \sqrt {{x^2} + {y^2}} y'\left( x \right) = 0 \cr
& {\text{multiplying}} \cr
& \frac{{xy}}{{\sqrt {{x^2} + {y^2}} }} + \frac{{{y^2}y'\left( x \right)}}{{\sqrt {{x^2} + {y^2}} }} + \sqrt {{x^2} + {y^2}} y'\left( x \right) = 0 \cr
& {\text{solving the equation for }}y'\left( x \right) \cr
& \frac{{{y^2}y'\left( x \right)}}{{\sqrt {{x^2} + {y^2}} }} + \sqrt {{x^2} + {y^2}} y'\left( x \right) = - \frac{{xy}}{{\sqrt {{x^2} + {y^2}} }} \cr
& \,\,\,\,\,\,\,\,\,\,\,{\text{factor }}y'\left( x \right) \cr
& \left( {\frac{{{y^2}}}{{\sqrt {{x^2} + {y^2}} }} + \sqrt {{x^2} + {y^2}} } \right)y'\left( x \right) = - \frac{{xy}}{{\sqrt {{x^2} + {y^2}} }} \cr
& \left( {\frac{{{y^2} + {x^2} + {y^2}}}{{\sqrt {{x^2} + {y^2}} }}} \right)y'\left( x \right) = - \frac{{xy}}{{\sqrt {{x^2} + {y^2}} }} \cr
& \left( {{x^2} + 2{y^2}} \right)y'\left( x \right) = - xy \cr
& y'\left( x \right) = - \frac{{xy}}{{{x^2} + 2{y^2}}} \cr} $$
