Answer
$$y = - \frac{{16}}{{23}}x + \frac{{94}}{{23}}$$
Work Step by Step
$$\eqalign{
& \sqrt {2y} - 4xy = - 22{\text{ at the point }}\left( {3,2} \right) \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {\sqrt {2y} - 4xy} \right) = \frac{d}{{dx}}\left( { - 22} \right) \cr
& \frac{d}{{dx}}\left( {\sqrt 2 \sqrt y } \right) - 4x\frac{d}{{dx}}\left( y \right) - y\frac{d}{{dx}}\left( {4x} \right) = \frac{d}{{dx}}\left( { - 22} \right) \cr
& {\text{solve the derivatives }} \cr
& \sqrt 2 \left( {\frac{1}{{2\sqrt y }}} \right)\frac{{dy}}{{dx}} - 4x\frac{{dy}}{{dx}} - y\left( 4 \right) = 0 \cr
& \frac{{\sqrt 2 }}{{2\sqrt y }}\frac{{dy}}{{dx}} - 4x\frac{{dy}}{{dx}} = 4y \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& \left( {\frac{{\sqrt 2 }}{{2\sqrt y }} - 4x} \right)\frac{{dy}}{{dx}} = 4y \cr
& \frac{{dy}}{{dx}} = \frac{{4y}}{{\frac{{\sqrt 2 }}{{2\sqrt y }} - 4x}} \cr
& {\text{find the slope at the given point }} \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {3,2} \right)}} = \frac{{4\left( 2 \right)}}{{\frac{{\sqrt 2 }}{{2\sqrt 2 }} - 4\left( 3 \right)}} = - \frac{{16}}{{23}} \cr
& {\text{find the equation of the tangent line at the point }}\left( {3,2} \right) \cr
& {\text{use the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 2 = - \frac{{16}}{{23}}\left( {x - 3} \right) \cr
& y - 2 = - \frac{{16}}{{23}}x + \frac{{48}}{{23}} \cr
& y = - \frac{{16}}{{23}}x + \frac{{94}}{{23}} \cr} $$
