Answer
$y=x+\frac{1}{2}$
Work Step by Step
$2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)\\ x+yy'=(2x^2+2y^2-x)(4x+4yy'-1)\\ x+yy'=8x^3+8x^2yy'-2x^2+8xy^2+8y^3y'-2y^2-4x^2-4xyy'+x\\ yy'-8x^2yy'-8y^3y'+4xyy'=8x^3-6x^2+8xy^2-2y^2\\ y'(y-8x^2y-8y^3+4xy)=8x^3-6x^2+8xy^2-2y^2\\ y'=\frac{8x^3-6x^2+8xy^2-2y^2}{y-8x^2y-8y^3+4xy}$
Plug in $(0,\frac{1}{2})$ into $y'$ to find the gradient
$y'=\frac{8(0)^3-6(0)^2+8(0)(\frac{1}{2})^2-2(\frac{1}{2})^2}{\frac{1}{2}-8(0)^2(\frac{1}{2})-8(\frac{1}{2})^3+4(0)(\frac{1}{2})}\\
y'=1$
Equation of the tangent line at $(0,\frac{1}{2})$
$y-\frac{1}{2}=1(x-0)\\
y=x+\frac{1}{2}$
