Find: $\lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital) 
Find: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)

After developing the expression, by multiplying the fraction by the conjugates and rearranging, I found:
$$f(x)=x\times \frac{1+\sqrt{1+1/x^6}}{1+\sqrt{1+1/x^4}}$$
I can solve it for the limit when $x\to \infty$, with answer $\infty$ which is correct (by the book and Wolfram Alpha). But when the limit is $x\to -\infty$ the answer is zero, but from this expression, I get $-\infty$ as the answer.
I'm certainly missing something. 
Hints and answers appreciated. Sorry if this is a duplicate.

Note that if $x\lt 0$, we have
$$\sqrt{x^6\left(1+\frac{1}{x^6}\right)}=|x^3|\sqrt{1+\frac{1}{x^6}}=\color{red}{-x^3}\sqrt{1+\frac{1}{x^6}}$$

$$\begin{align}&\lim_{x\to -\infty}\frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}\\\\&=\lim_{x\to -\infty}\frac{(x^2-\sqrt{x^4+1})(x^2+\sqrt{x^4+1})(x^3+\sqrt{x^6+1})}{(x^3-\sqrt{x^6+1})(x^2+\sqrt{x^4+1})(x^3+\sqrt{x^6+1})}\\\\&=\lim_{x\to -\infty}\frac{x^3+\sqrt{x^6+1}}{x^2+\sqrt{x^4+1}}\\\\&=\lim_{x\to -\infty}\frac{x^3+\sqrt{x^6(1+\frac {1}{x^6})}}{x^2+\sqrt{x^4(1+\frac{1}{x^4})}}\\\\&=\lim_{x\to -\infty}\frac{x^3+|x^3|\sqrt{1+\frac{1}{x^6}}}{x^2+|x^2|\sqrt{1+\frac{1}{x^4}}}\\\\&=\lim_{x\to -\infty}\frac{x^3+(\color{red}{-x^3})\sqrt{1+\frac{1}{x^6}}}{x^2+x^2\sqrt{1+\frac{1}{x^4}}}\\\\&=\lim_{x\to -\infty}x\cdot\frac{\left(1-\sqrt{1+\frac{1}{x^6}}\right)\left(1+\sqrt{1+\frac{1}{x^6}}\right)}{\left(1+\sqrt{1+\frac{1}{x^4}}\right)\left(1+\sqrt{1+\frac{1}{x^6}}\right)}\\\\&=\lim_{x\to -\infty}\frac{-1}{x^5\left(1+\sqrt{1+\frac{1}{x^4}}\right)\left(1+\sqrt{1+\frac{1}{x^6}}\right)}=0\end{align}$$
