Answer
$$y = {\left( {\frac{3}{2}\tan x - \frac{3}{2}x + \sqrt 3 } \right)^2}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = 3\sqrt y {\tan ^2}x,{\text{ }}\left( {0,3} \right) \cr
& {\text{Separate the variables}} \cr
& {y^{ - 1/2}}dy = 3{\tan ^2}xdx \cr
& {\text{Integrate both sides}} \cr
& \int {{y^{ - 1/2}}} dy = 3\int {\left( {{{\sec }^2}x - 1} \right)} dx \cr
& 2\sqrt y = 3\tan x - 3x + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {0,3} \right) \cr
& 2\sqrt 3 = 3\tan \left( 0 \right) - 3\left( 0 \right) + C \cr
& C = 2\sqrt 3 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& 2\sqrt y = 3\tan x - 3x + 2\sqrt 3 \cr
& \sqrt y = \frac{3}{2}\tan x - \frac{3}{2}x + \sqrt 3 \cr
& y = {\left( {\frac{3}{2}\tan x - \frac{3}{2}x + \sqrt 3 } \right)^2} \cr
& \cr
& {\text{Graph}} \cr} $$
