Question

Dy/dx = y/x^2

dy/y = dx/x^2

ln y * ln c = -1/x

cy = e^(-1/x)

y = ce^(-1/x)

Answer 1

dy/dx = y/(x^2)
dy/y = dx/(x^2)
int[dy/y] = int[dx/(x^2)] ... apply integral to both sides
ln(|y|) = (-1/x) + C
|y| = e^{(-1/x) + C}
|y| = e^C*e^(-1/x)
|y| = C*e^(-1/x)
y = C*e^(-1/x)

So you have the correct answer. Nice job.

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Check:
y = C*e^(-1/x)
dy/dx = d/dx[C*e^(-1/x)]
dy/dx = d/dx[-1/x]*C*e^(-1/x)
dy/dx = (1/(x^2))*C*e^(-1/x)
is the expression for the left hand side (LHS)

y/(x^2) = [C*e^(-1/x)]/(x^2)
y/(x^2) = (1/(x^2))*C*e^(-1/x)
is the expression for the right hand side (RHS)

Since LHS = RHS, this confirms the solution for dy/dx = y/(x^2)