Let $k$ be a field and $k[x,y,z]$ be a polynomial ring in the indeterminates $x,y,z$. Define the quotient ring $$R = k[x,y,z]/(xz - (y^2 + 1)).$$
I guess I was able to show that $R$ is an integral domain. Now, I want to find the field of fractions $F = \operatorname{frac}(R).$ I know that it should be $R(x,y,z)$ (is that correct?), but I am very confused how to find it exactly.
Any help will be appreciated!
EDIT
I knew that the field of fraction of $R$ is $k(x,y)$ or $k(y,z)$ whether you inverted $x$ or $z.$ But now what I am really stuck in the step of the proof that $k(y,z) \subseteq F.$ What exactly should I write to show this inclusion?
Any help is greatly appreciated!
Let me flesh out the comments of runway44 under your question. Let $\varphi:k[y,z]\to R$ be the canonical embedding given by $y\mapsto\overline{y}$ and $z\mapsto\overline{z}$. (If you prefer, $\varphi$ is the composition of the inclusion $k[y,z]\hookrightarrow k[x,y,z]$ with the projection map $k[x,y,z]\to R$.) Note that $\varphi$ is injective; indeed, suppose some $p(y,z)\in k[y,z]$ is mapped by $\varphi$ to $\overline{0}\in R$. Then in particular $p(y,z)$, considered as an element of $k[x,y,z]$, lies in the ideal $P:=\langle xz-(y^2+1)\rangle\triangleleft k[x,y,z]$. But any non-zero element of $P$ will have monomial terms in which $x$ appears non-trivially (why?), and $p(y,z)$ does not have any monomial terms in which $x$ appears non-trivially, so we must have $p(y,z)=0$ as desired.
Now, let $Q(R)$ denote the field of fractions of $R$, and let $i:R\hookrightarrow Q(R)$ be the canonical inclusion. Then $i\circ\varphi$ is an injection from the domain $k[y,z]$ into the field $Q(R)$, and so, by the universal property of fields of fractions, $i\circ\varphi$ extends to an injection $\psi:k(y,z)\hookrightarrow Q(R)$. We claim that $\psi$ is surjective, and hence an isomorphism, which will show $Q(R)\cong k(y,z)$.
First we claim it suffices to show that $\overline{x}\in Q(R)$ lies in the image of $\psi$. Why? Well, suppose there is some element $a\in k(y,z)$ such that $\psi(a)=\overline{x}$. Now, note that an arbitrary element of $Q(R)$ is of the form $f\big/g$ for some $f\in R$ and some $g\in R\setminus\{\overline{0}\}$. Choose representative polynomials $p,q\in k[x,y,z]$ with $p(\overline{x},\overline{y},\overline{z})=f$ and $q(\overline{x},\overline{y},\overline{z})=g$ in $R$. Then $q(a,y,z)$ is non-zero in $k(y,z)$; indeed, otherwise we would have $$g=q(\overline{x},\overline{y},\overline{z})=q(\psi(a),\psi(y),\psi(z))=\psi(q(a,y,z))=\psi(0)=\overline{0},$$ a contradiction. In particular, the fraction $p(a,y,z)\big/q(a,y,z)$ is a well-defined element of $k(y,z)$, and we have $$\psi\left(\frac{p(a,y,z)}{q(a,y,z)}\right)=\frac{p(\psi(a),\psi(y),\psi(z))}{q(\psi(a),\psi(y),\psi(z))}=\frac{p(\overline{x},\overline{y},\overline{z})}{q(\overline{x},\overline{y},\overline{z})}=\frac{f}{g}.$$ Since $f\big/g$ was arbitrary, this shows that $\psi$ is surjective, as desired.
So, we need only show that $\overline{x}\in Q(R)$ lies in the image of $\psi$. To see this, consider the element $(y^2+1)\big/z\in k(y,z)$. Its image under $\psi$ will be $(\overline{y}^2+\overline{1})\big/\overline{z}$. But we know $\overline{x}\cdot\overline{z}=\overline{y}^2+\overline{1}$ in $R$ (why?), so dividing out by the unit $\overline{z}\in Q(R)$ gives $\overline{x}=(\overline{y}^2+\overline{1})\big/\overline{z}$ in $Q(R)$. In particular, we have $\psi\left((y^2+1)\big/z\right)=\overline{x}$, and so by the paragraph above we are done.