We identify that the differential equation $\frac{dy}{dx}+\frac{-y}{x}=\frac{x}{3y}$ is a Bernoulli differential equation since it's of the form $\frac{dy}{dx}+P(x)y=Q(x)y^n$, where $n$ is any real number different from $0$ and $1$. To solve this equation, we can apply the following substitution. Let's define a new variable $u$ and set it equal to

Plug in the value of $n$, which equals $-1$

Simplify

Now, substitute $\frac{dy}{dx}=\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}$ and $y=\sqrt{u}$ on the original differential equation

Simplify

We need to cancel the term that is in front of $\frac{du}{dx}$. We can do that by multiplying the whole differential equation by $\frac{1}{2}\sqrt{u}$

Multiply both sides by $\frac{1}{2}\sqrt{u}$

We can identify that the differential equation has the form: $\frac{dy}{dx} + P(x)\cdot y(x) = Q(x)$, so we can classify it as a linear first order differential equation, where $P(x)=\frac{-1}{\frac{1}{2}x}$ and $Q(x)=\frac{2x}{3}$. In order to solve the differential equation, the first step is to find the integrating factor $\mu(x)$

We can recognize that the left side of the differential equation consists of the derivative of the product of $\mu(x)\cdot y(x)$

Integrate both sides of the differential equation with respect to $dx$

Simplify the left side of the differential equation

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

Any expression to the power of $1$ is equal to that same expression

Replace $u$ with the value $y^{2}$