Answer
$$ - \frac{{24}}{{13}}$$
Work Step by Step
$$\eqalign{
& x\root 3 \of y + y = 10,{\text{ }}\left( {1,8} \right) \cr
& x{y^{1/3}} + y = 10 \cr
& {\text{Differentiate both sides with respect to }}x \cr
& x\left( {\frac{1}{3}{y^{ - 2/3}}} \right)y' + {y^{1/3}} + y' = 0 \cr
& \frac{1}{3}x{y^{ - 2/3}}y' + {y^{1/3}} + y' = 0 \cr
& \frac{1}{3}x{y^{ - 2/3}}y' + y' = - {y^{1/3}} \cr
& \left( {\frac{1}{3}x{y^{ - 2/3}} + 1} \right)y' = - {y^{1/3}} \cr
& y' = \frac{{ - {y^{1/3}}}}{{\frac{1}{3}x{y^{ - 2/3}} + 1}} \cr
& {\text{Evaluate at the point }}\left( {1,8} \right) \cr
& y' = \frac{{ - {{\left( 8 \right)}^{1/3}}}}{{\frac{1}{3}\left( 1 \right){{\left( 8 \right)}^{ - 2/3}} + 1}} \cr
& y' = \frac{{ - 2}}{{13/12}} \cr
& y' = - \frac{{24}}{{13}} \cr
& {\text{The slope at the given point is }} - \frac{{24}}{{13.}} \cr} $$
