Answer
$$y = - \frac{4}{5}x + \frac{{24}}{5}$$
Work Step by Step
$$\eqalign{
& y + \sqrt {xy} = 6{\text{ point }}\left( {1,4} \right) \cr
& {\text{use implicit differentiation}}{\text{, differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ y \right] + \frac{d}{{dx}}\left[ {\sqrt x \sqrt y } \right] = \frac{d}{{dx}}\left[ 6 \right] \cr
& {\text{using the product rule}} \cr
& \frac{d}{{dx}}\left[ y \right] + \sqrt x \frac{d}{{dx}}\left[ {\sqrt y } \right] + \sqrt y \frac{d}{{dx}}\left[ {\sqrt x } \right] = \frac{d}{{dx}}\left[ 6 \right] \cr
& {\text{solve derivatives}} \cr
& y'\left( x \right) + \sqrt x \left( {\frac{1}{{2\sqrt y }}} \right)y'\left( x \right) + \sqrt y \left( {\frac{1}{{2\sqrt x }}} \right) = 0 \cr
& {\text{simplify}} \cr
& y'\left( x \right) + \frac{1}{2}\sqrt {\frac{x}{y}} y'\left( x \right) + \frac{1}{2}\sqrt {\frac{y}{x}} = 0 \cr
& {\text{Solve for }}y'\left( x \right) \cr
& \left( {1 + \frac{1}{2}\sqrt {\frac{x}{y}} } \right)y'\left( x \right) = - \frac{1}{2}\sqrt {\frac{y}{x}} \cr
& \,\,y'\left( x \right) = \frac{{ - \frac{1}{2}\sqrt {\frac{y}{x}} }}{{1 + \frac{1}{2}\sqrt {\frac{x}{y}} }} \cr
& {\text{find the slope at the point }}\left( {1,4} \right) \cr
& \,\,\,m = \frac{{ - \frac{1}{2}\sqrt 4 }}{{1 + \frac{1}{2}\sqrt {1/4} }} \cr
& \,\,\,m = - \frac{4}{5} \cr
& {\text{find the equation of the tangent line at the point }}\left( {1,4} \right) \cr
& \,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,\,\,\,y - 4 = - \frac{4}{5}\left( {x - 1} \right) \cr
& {\text{simplify}} \cr
& \,\,\,\,\,y - 4 = - \frac{4}{5}x + \frac{4}{5} \cr
& \,\,\,\,\,y = - \frac{4}{5}x + \frac{4}{5} + 4 \cr
& \,\,\,\,\,y = - \frac{4}{5}x + \frac{{24}}{5} \cr} $$
