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?

The step "In order for (1) to hold..." is incorrect. Just before it, you had an expression of the form $A-B=0$ ($A$ happens to be $x_2^2(x_1^3+9)$ and $B$ happens to be $x_1^2(x_2^3+9)$.) From it you have concluded that $A=0$ and $B=0$. This is not correct - the difference of two numbers can be $0$ even if neither of them is zero (as long as they are equal, that is!).
Now, maybe it will help to write the original function as:
$$g(x)=x+\frac{9}{x^2}$$
You can see that both addends here are strictly increasing. Let's assume $x_1<x_2<0$, you have $x_1^2>x_2^2$ and so $\frac{9}{x_1^2}<\frac{9}{x_2^2}$, and finally, $g(x_1)=x_1+\frac{9}{x_1^2}\lt x_2+\frac{9}{x_2^2}=g(x_2)$
Thus $g(x)$ on $(-\infty, 0)$ is "1-1".
