Is $g(x) = \frac{x^3+9}{x^2}$ one-to-one? Let $g(x) =\frac{x^3+9}{x^2}$ restricted to $D(g) = (-\infty,0)$. Is it 1-1?

My approach
$g(x) \text{ 1-1 }: g(x_1)=g(x_2) \iff x_1 = x_2$
Let $x_1, x_2 \in D(g)$ then,
$$\frac{x_1^3+9}{x_1^2} = \frac{x_2^3+9}{x_2^2} \iff x_2^2 \cdot x_1^3 + 9x_2^2 -  x_1^2 \cdot x_2^3 + 9x_1^2 =0 \iff\\ 
\iff x_2^2(x_1^3+9)-x_1^2(x_2^3+9) = 0 \quad (1)$$
In order for $(1)$ to hold,

*

*$x_2 = x_1 = 0$, which can't be because $0 \notin D(g)$ or

*$x_1 = x_2 = \sqrt[3]{-9} \in D(g)$
Therefore $g(x_1)=g(x_2) \iff x_1 = x_2$ holds only for $\sqrt[3]{-9}$ (and not for every $x \in D(g)$)
$$\boxed{ \text{Hence, }g(x)\text{ is NOT 1-1  }}$$
Is this correct?

I would use a calculus approach:
$$g(x) = \frac{x^3+9}{x^2}=x+\frac{9}{x^2}$$
Differentiating gives us $$1-\frac{9}{x^3}>0$$ which means it is a monotonic function over the negative domain.
