Implicit Integral

$\begin{aligned} \frac {dy}{dx} & = \frac 3{\frac xy+2} - 2 & \small {\color{#3D99F6}\text{Let }y=vx} \\ v + x \frac {dv}{dx} & = \frac 1{\frac 1v+2} - 2 \\ x \frac {dv}{dx} & = - \frac {2(v^2+v+1)}{2v+1} \\ \frac {2v+1}{v^2+v+1} dv & = - \frac 2x dx & \small {\color{#3D99F6}\text{Integrate both sides}} \\ \ln (v^2+v+1) & = - 2\ln x \\ v^2+v+1 & = \frac 1{x^2} \\ \frac {y^2}{x^2} + \frac yx + 1 & = \frac 1{x^2} \\ y^2+xy+x^2 & = {\color{#3D99F6}C} & \small {\color{#3D99F6}\text{where }C \text{ is the constant of integration.}} \\ \implies C & = 3 & \small {\color{#3D99F6}\text{Putting }x=1 \text{ and }y=1} \\ \implies y^2+xy+x^2 & = C \quad ...(1) \end{aligned}$

Maximum of $y$ occurs when $\dfrac {dy}{dx} = 0$ , that is:

$\begin{aligned} \frac 3{\frac xy+2} & = 2 \\ \frac {3y}{x+2y} & = 2 \\ \implies y & = - 2x \\ (1): \quad 4x^2-2x^2+x^2 & = 3 \\ 3x^2 & = 3 \\ x & = \pm 1 \end{aligned}$

Therefore, maximum $y_{max} = -2(-1) = \boxed{2}$ as $\dfrac {d^2y}{dx^2}\bigg|_{x=-1} < 0$ (see Note).

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Note:

$\begin{aligned} \frac {dy}{dx} & = \frac 3{\frac xy+2} - 2 = -\frac {2x+y}{x+2y} \\ \frac {d^2y}{dx^2} & = -\frac {\left(2+\frac {dy}{dx}\right)(x+2y)+\left(1+2\frac {dy}{dx}\right)(2x+y)}{(x+2y)^2} & \small {\color{#3D99F6}\text{Putting }x=-1} \\ & = -\frac {\left(2+0\right)(-1+4)+\left(1+0\right)(-2+2)}{(-1+4)^2} \\ & < 0 \end{aligned}$

After simplifying the equation :

$\displaystyle \frac{dy}{dx}=\frac{-2x-y}{x+2y} \\ \implies xdy+2ydy=-2xdx-ydx \\ \implies xdy+ydx+2ydy+2xdx=0 \\ \implies d(xy)+d(x^2+y^2)=0 \\ \color{#3D99F6}\text{Integrating both sides} \\ \implies xy+x^2+y^2=c \:\: \text{where c is an integration constant} \\ \text{putting x=1 and y=1 in the equation the c=3} \\ \implies x^2+y^2+xy=3 \\ \text{Using quadratic w.r.t. x }\\ D=\sqrt{y^2-4(y^2-3)}\geq 0 \\ 3y^2\leq12 \\ \implies |y|\leq2 \\ \text{So, maximum value of y=2}$