x^3+x^2y+2y^2=0

what is the question?

since the topic is calculus, I guess they want to show that

y' = -(3x^2+2xy)/(x^2+4y)

The given equation is a quadratic equation in terms of variables x and y. To find the solutions (values of x and y) that satisfy the equation, we can use a technique called factoring.

Step 1: Rewrite the equation in a more manageable form if possible. In this case, we don't have a common factor between the terms, so we can skip this step.

Step 2: Look for opportunities to factor the equation. In this case, the equation is already in factored form. We can see that all three terms have a common factor of x, so we can factor out an x: x(x^2 + xy + 2y^2) = 0.

Step 3: Set each factor equal to zero. Since a product of factors is zero if and only if at least one of the factors is zero, we have two equations to solve: x = 0 and x^2 + xy + 2y^2 = 0.

For x = 0, the entire equation is satisfied.

For x^2 + xy + 2y^2 = 0, this equation represents a quadratic equation in terms of y. To solve this quadratic equation, we can either use factoring, completing the square, or the quadratic formula. Unfortunately, this quadratic equation is not easily factorable, so we will use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac))/(2a)

Comparing our equation to the quadratic formula, we can see that a = 1, b = 1, and c = 2. Substituting these values into the quadratic formula, we get:

x = (-1 ± √(1^2 - 4(1)(2)))/(2(1))

Simplifying further, we have:

x = (-1 ± √(1 - 8))/2
x = (-1 ± √(-7))/2

Since the square root of a negative number is not a real number, we can conclude that there are no real solutions for x^2 + xy + 2y^2 = 0.

Thus, the solutions for the equation x^3 + x^2y + 2y^2 = 0 are x = 0 and there are no real solutions for x^2 + xy + 2y^2 = 0.