What is the value of $f(1999)$? Let be a function $f \colon \Bbb R\to \Bbb R$ given from:
$$f(x)=x^3+\sqrt{x^6+1}+\frac{1}{x^3-\sqrt{x^6+1}}$$
What is the value of $f(1999)$?
To me, it immediately seemed strange that it was necessary to calculate $f(1999)$. With a calculator everything would come easy. Since this question concerns high school students I have done the rationalization of $\frac{1}{x^3-\sqrt{x^6+1}}$ and I have found something very interesting. In fact
$$\frac{1}{x^3-\sqrt{x^6+1}}=\frac{1}{x^3-\sqrt{x^6+1}}\cdot\frac{x^3+\sqrt{x^6+1}}{x^3+\sqrt{x^6+1}}=\frac{x^3+\sqrt{x^6+1}}{-1}$$
because $(x^3-\sqrt{x^6+1})(x^3+\sqrt{x^6+1})=-1$
Thus
$$f(x)=x^3+\sqrt{x^6+1}-x^3-\sqrt{x^6+1}=0$$
and $f(1999)=0$ because $f(x)=0, \forall x\in\Bbb R$. Is it correct?

Your work seems correct to me. (this is stated in the comments that your work is correct)
The problem statement is essentially equivalent to:

Let $f:\mathbb R\rightarrow \mathbb R$,
$$f(x):=A(x)+B(x)+\frac {1}{A(x)-B(x)}$$
where, $A^2(x)-B^2(x)=-1$.
Find the value of $f(1999).$

$$
\begin{align}f(x):&=A(x)+B(x)+\frac{A(x)+B(x)}{A^2(x)-B^2(x)}\\
&=A(x)+B(x)-\left(A(x)+B(x)\right)\\
&=0~.\end{align}
$$
Thus $f(x)\equiv 0$, $\forall x\in\mathbb R$, since $\operatorname{dom}f=\mathbb R.$
What we do is based on using conjugate.
Note that, when you add fractions, the conjugate will appear by itself. Indeed,
$$
\begin{align}f(x):&=A(x)+B(x)+\frac {1}{A(x)-B(x)}\\
&=\frac {\left(A(x)-B(x)\right)\cdot\left(A(x)+B(x)\right)+1}{A(x)-B(x)}\\
&=\frac{\overbrace{A^2(x)-B^2(x)}^{\color {#c00}{-1}}+1}{A(x)-B(x)}\\
&=0~.
\end{align}
$$
Because, $A^2(x)-B^2(x)=-1$ implies that, $A(x)±B(x)≠0$.
Finally, I want to note that this type of function is called a constant function in mathematics.
