I am having trouble with the following qualifying exam problem and I would appreciate any help. Thank you.
Let $X$ be the set of pairs of unit vectors $(x, y)$ in $\mathbb{R}^3$ such that $x \cdot y = \frac{1}{2}$ where $\cdot$ denotes the dot product. Thus $X = \left \{(x, y) \in \mathbb{R}^3 \times \mathbb{R}^3 \ | \ |x| = 1, |y| = 1, x\cdot y = \frac{1}{2}\right\}$. Prove that $X$ is a manifold.
Consider the function $f:\mathbb R^3\times \mathbb R^3\to \mathbb R^3$ defined by $$f(x_1,x_2,x_3,y_1,y_2,y_3)=(x_1^2+x_2^2+x_3^2,y_1^2+y_2^2+y_3^2,x_1y_1+x_2y_2+x_3y_3)$$ and note that $X=f^{-1}(1,1,1/2)$. Clearly $f$ is smooth, so in order to show that $f^{-1}(1,1,1/2)$ is a manifold, it suffices to show that $(1,1,1/2)$ is a regular value of $f$. The derivative of $f$ is $$\begin{pmatrix} 2x_1 & 2x_2 & 2x_3 & 0 & 0 & 0\\ 0 & 0 & 0 & 2y_1 & 2y_2 & 2y_3\\ y_1 & y_2 & y_3 & x_1 & x_2 & x_3\\ \end{pmatrix}$$ and we want to show that this is surjective when $f(x_1,x_2,x_3,y_1,y_2,y_3)=(1,1,1/2)$. Equivalently, we want to show that its row-rank is $3$, i.e. the rows are linearly independent. Clearly the first and second are linearly independent since we have some $x_i$ and $y_i$ which are nonzero. Thus the only linear dependence possible has nonzero coefficient on the last row, so we can take it to be $2$, thus we have some $a,b\in\mathbb R$ such that $x_i=ay_i$ and $y_i=bx_i$ for all $i$ (and $a=1/b$). But then $1/2=x\cdot y = x\cdot (bx)=b|x|=b$, and similarly $1/2=a$, contradicting the fact that $a=1/b$. Thus the derivative is surjective, so $X$ is a manifold.