We are given the first-order differential equation:

\[
x(x + y)y' + y(3x + y) = 0
\]

Our goal is to solve this equation.

### Step 1: Rearrange the equation
First, divide the entire equation by \( x \), assuming \( x \neq 0 \):

\[
(x + y)y' + \frac{y(3x + y)}{x} = 0
\]

### Step 2: Simplify and inspect
Rewrite the terms more clearly:

\[
(x + y)y' + y \left(\frac{3x}{x} + \frac{y}{x}\right) = 0
\]
\[
(x + y)y' + y(3 + \frac{y}{x}) = 0
\]

### Step 3: Change of variables
To reduce the complexity, we can try the substitution \( v = \frac{y}{x} \), which implies \( y = vx \). Differentiate both sides with respect to \( x \):

\[
y' = v + x \frac{dv}{dx}
\]

Substitute \( y = vx \) and \( y' = v + x \frac{dv}{dx} \) into the equation:

\[
(x + vx)\left(v + x \frac{dv}{dx}\right) + vx(3 + v) = 0
\]

### Step 4: Simplify the equation
Expand the terms:

\[
x(v + 1)\left(v + x \frac{dv}{dx}\right) + vx(3 + v) = 0
\]
\[
x(v + 1)v + x(v + 1)x \frac{dv}{dx} + vx(3 + v) = 0
\]
\[
x(v + 1)x \frac{dv}{dx} + x \left( (v + 1)v + v(3 + v) \right) = 0
\]
Now simplify the second part:
\[
(v + 1)v + v(3 + v) = v^2 + v + 3v + v^2 = 2v^2 + 4v
\]
Thus, the equation becomes:

\[
x(v + 1)x \frac{dv}{dx} + x(2v^2 + 4v) = 0
\]
\[
x^2(v + 1) \frac{dv}{dx} + x(2v^2 + 4v) = 0
\]

### Step 5: Separate variables
We can divide through by \( x \), assuming \( x \neq 0 \):

\[
x(v + 1) \frac{dv}{dx} + (2v^2 + 4v) = 0
\]

Rearrange it:

\[
x(v + 1) \frac{dv}{dx} = -(2v^2 + 4v)
\]

Now divide both sides by \( v + 1 \):

\[
x \frac{dv}{dx} = -\frac{2v^2 + 4v}{v + 1}
\]

### Step 6: Simplify the right-hand side
Simplify the expression on the right:

\[
\frac{2v^2 + 4v}{v + 1} = 2v
\]

So the equation reduces to:

\[
x \frac{dv}{dx} = -2v
\]

### Step 7: Solve the differential equation
Separate the variables \( v \) and \( x \):

\[
\frac{dv}{v} = -\frac{2}{x} dx
\]

Integrate both sides:

\[
\ln |v| = -2 \ln |x| + C_1
\]

Simplify the right-hand side:

\[
\ln |v| = \ln \left(\frac{C_2}{x^2}\right)
\]

Exponentiate both sides:

\[
|v| = \frac{C_2}{x^2}
\]

Thus:

\[
v = \frac{C_2}{x^2}
\]

Since \( v = \frac{y}{x} \), we have:

\[
\frac{y}{x} = \frac{C_2}{x^2}
\]

### Step 8: Solve for \( y \)
Multiply both sides by \( x \):

\[
y = \frac{C_2}{x}
\]

Thus, the general solution to the differential equation is:

\[
y = \frac{C}{x}
\]
where \( C = C_2 \) is a constant.

---

Would you like more details on any part of the solution?

### Related Questions:
1. How does substitution help in solving differential equations?
2. What are the general steps to solve separable differential equations?
3. How does the assumption \( x \neq 0 \) affect the solution?
4. Could a different substitution method be used here?
5. What are other common methods to solve first-order differential equations?

### Tip:  
In solving differential equations, substitution is a powerful tool for simplifying complex expressions and reducing the equation to a form that can be more easily integrated.

This tutorial provides a step-by-step solution to the first-order differential equation x(x + y)y' + y(3x + y) = 0. It covers essential concepts like substitution and separation of variables, making it ideal for advanced high school or undergraduate students.

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