If \((x+2y)\cdot \dfrac{dy}{dx}=2x-y\), what is the value of \(\dfrac{d^2y}{dx^2}\) at the point \((3,0)\) ?

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$$ (x+2y)\cdot \dfrac{dy}{dx}=2x-y $$

Solve for \(\frac{dy}{dx}\) and differentiate.

$$ \frac{dy}{dx}=\frac{2x-y}{x+2y} $$ $$ \frac{d^2y}{dx^2}=\frac{(x+2y)(2-\frac{dy}{dx})-(2x-y)(1+2\frac{dy}{dx})}{(x+2y)^2} $$

Find the value of \(\frac{dy}{dx}\) at the point:

$$ \frac{dy}{dx}\Big|_{(3,0)}=\frac{2(3)-0}{3+2(0)} $$ $$\frac{dy}{dx}\Big|_{(3,0)} = 2 $$

Substitute known values into the second derivative:

$$ \frac{d^2y}{dx^2}=\frac{((3)+2(0))(2-(2))-(2(3)-(0))(1+2(2))}{((3+2(0))^2} $$ $$ = \frac{(3)(0)-(6)(5)}{3^2} $$ $$ = -\frac{30}{9} $$ $$ = \boxed{-\frac{10}{3}} $$