What is the implicit derivative of -2=x (x+y)^2 -3xy?

$y ' = \frac{3 {x}^{2} + 4 x y - 3 y + {y}^{2}}{3 x - 2 x y - 2 {x}^{2}}$

Explanation:

From the given equation $- 2 = x {\left(x + y\right)}^{2} - 3 x y$ differentiate both sides with respect to x

$- 2 = x {\left(x + y\right)}^{2} - 3 x y$

$\frac{d}{\mathrm{dx}} \left(- 2\right) = \frac{d}{\mathrm{dx}} \left[x {\left(x + y\right)}^{2} - 3 x y\right]$

$0 = x \cdot \frac{d}{\mathrm{dx}} \left({\left(x + y\right)}^{2}\right) + {\left(x + y\right)}^{2} \cdot \frac{d}{\mathrm{dx}} \left(x\right) - 3 \cdot \left[x \cdot \frac{d}{\mathrm{dx}} \left(y\right) + y \cdot \frac{d}{\mathrm{dx}} \left(x\right)\right]$

$0 = x \cdot 2 {\left(x + y\right)}^{2 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x + y\right) + {\left(x + y\right)}^{2} \cdot 1 - 3 \left[x \cdot y ' + y \cdot 1\right]$

$0 = 2 x \cdot \left(x + y\right) \cdot \left(1 + y '\right) + {\left(x + y\right)}^{2} - 3 x y ' - 3 y$

$0 = \left(2 {x}^{2} + 2 x y\right) \cdot \left(1 + y '\right) + {x}^{2} + 2 x y + {y}^{2} - 3 x y ' - 3 y$

$0 = 2 {x}^{2} + 2 x y + 2 {x}^{2} y ' + 2 x y y ' + {x}^{2} + 2 x y + {y}^{2} - 3 x y ' - 3 y$

transpose all the terms with $y '$ to the left side of the equation

$3 x y ' - 2 x y y ' - 2 {x}^{2} y ' = 2 {x}^{2} + 2 x y + {x}^{2} + 2 x y + {y}^{2} - 3 y$

simplify

$3 x y ' - 2 x y y ' - 2 {x}^{2} y ' = 3 {x}^{2} + 4 x y - 3 y + {y}^{2}$

factor out the common monomial factor $y '$

$\left(3 x - 2 x y - 2 {x}^{2}\right) y ' = 3 {x}^{2} + 4 x y - 3 y + {y}^{2}$

Divide by the coefficient of $y '$

$\frac{\left(3 x - 2 x y - 2 {x}^{2}\right) y '}{3 x - 2 x y - 2 {x}^{2}} = \frac{3 {x}^{2} + 4 x y - 3 y + {y}^{2}}{3 x - 2 x y - 2 {x}^{2}}$

$\frac{\cancel{3 x - 2 x y - 2 {x}^{2}} y '}{\cancel{3 x - 2 x y - 2 {x}^{2}}} = \frac{3 {x}^{2} + 4 x y - 3 y + {y}^{2}}{3 x - 2 x y - 2 {x}^{2}}$

$y ' = \frac{3 {x}^{2} + 4 x y - 3 y + {y}^{2}}{3 x - 2 x y - 2 {x}^{2}}$

God bless....I hope the explanation is useful.