Given,

$${{dy} \over {dx}} + {y \over 2}\sec x = {{\tan x} \over {2y}}$$

$$ \Rightarrow $$   $$2y{{dy} \over {dx}} + {y^2}\sec x = \tan x$$

Now, let

y² $$=$$ t

$$ \Rightarrow $$   2y$${{dy} \over {dx}} = {{dt} \over {dx}}$$

$$ \therefore $$    New equation,

$${{dt} \over {dx}} + t\sec x = \tan x$$

$$ \therefore $$   I.F $$=$$ $${e^{\int {\sec xdx} }}$$

$$=$$   $${e^{\ln \left( {\sec x + \tan x} \right)}}$$

$$=$$   sec x + tan x

$$ \therefore $$   Solution is,

t(sec x + tan x) $$=$$ $$\int {\tan x} $$ (sec x + tan x) dx

$$ \Rightarrow $$   t(sec x + tan x) $$=$$ sec x + tan x $$-$$ x + c

$$ \Rightarrow $$    t $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}} + c$$

$$ \Rightarrow $$   y² $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}} + c$$

Given,

y(0) $$=$$ 1

$$ \therefore $$   1 $$=$$ 1 $$-$$ 0 + c

$$ \Rightarrow $$   c $$=$$ 0

$$ \therefore $$   y² $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}}$$