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:

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