The following example is elementary and you can find the main argument in Klaus Hulek's ''Elementare algebraische Geometrie'' in the introduction. I add some details.
Fix a complex number $\lambda \in \mathbb{C} \setminus \{0,1\}$. Define $$f(x,y,z) = y^2 z -x(x-z)(x- \lambda z)$$ and let $X := V(f) \subset \mathbb{P}^2$. I claim that $X$ is not rational.
Let $$g(x,y) := f(x,y,1) = y^2 - x(x-1)(x- \lambda).$$ Let $U_z := \{[x:y:z] \in \mathbb{P}^2 \, : \, z \neq 0\}$ and let $\varphi : U_z \rightarrow \mathbb{C}^2$ with $\varphi([x:y:z]) =(x/z,y/z)$. Then $\varphi(U_z \cap X) = V(g) =: Y$ and hence $U_z \cap X$ is isomorphic to $Y$, so that $Y$ is birational to $X$.
You can check that $Y$ is irreducible and nonsingular (here we use $\lambda \neq 0,1$), so the same holds for $X \cap U_z$. We have $X \cap (\mathbb{P}^2\setminus U_z) = [0:1:0] \in U_y$ and one can check that $[0:1:0]$ is not a singular point of $X$. It follows that $X$ is a nonsingular irreducible projective curve, which is birational to $Y$ and hence it suffices to show that $Y$ is not rational.
Let $K:= Q(\mathbb{C}[x,y]/(g))$ be the field of rational functions on $Y$ (where $Q(A)$ denotes the field of fractions of an integral domain $A$). Suppose $Y$ were rational. Then there would be a field isomorphism $\Phi: K \rightarrow \mathbb{C}(t)$. Let $\varphi(t) := \Phi(x)$ and $\psi(t):= \Phi(y)$ in $\mathbb{C}(t)$ (where instead of $x$ we could pedantically write $\frac{x+(f)}{1+(f)}$). Note that not both of $\varphi(t)$ and $\psi(t)$ can be constant, as this would contradict the surjectivity of $\Phi$. We have the identity $$ 0=\Phi(g(x,y)) = g(\Phi(x), \Phi(y)) = g(\varphi(t), \psi(t)) $$ in $\mathbb{C}(t)$. So we will obtain a contradiction by showing that whenever we have rational functions $\varphi(t)$ and $\psi(t)$ satisfying $g(\varphi(t), \psi(t)) = 0$, one of them has to be constant.
Write $\varphi = p/q$ and $\psi = r/s$ with coprime polynomials $p,q \in \mathbb{C}[t]$ and coprime polynomials $r,s \in \mathbb{C}[t]$. Then
$$ (r/s)^2 = (p/q)((p/q) -1)((p/q)- \lambda)$$
or
$$r^2q^3 = s^2p(p-q)(p-\lambda q) \in \mathbb{C}[t]$$
Now $s^2$ divides the LHS, but not $r^2$ hence $s^2$ divides $q^3$. Similalry, $q^3$ divides the RHS, but not $p(p-q)(p - \lambda q)$ hence $q^3$ divides $s^2$. This implies that $a s^2 = q^3$ for some $a \in \mathbb{C}^\times$. Hence $q$ is a square in $\mathbb{C}[t]$. Replacing $q^3$ with $as^2$ in the displayed equation we see that
$$ar^2 = p(p-q)(p- \lambda q)$$
The polynomials $p$, $p-q$ and $p - \lambda q$ are coprime to each other but their product is a square. Consequently, all of $p$, $p-q$ and $p- \lambda q$ have to be squares in $\mathbb{C}[t]$ (and also $q$ as noted before). We may now invoke Lemma 0.8 on page 7 of Hulek's book ''Elementare algebraische Geometrie'' (where the heart of this answer is taken from). I would like to repharse it slightly as follows.
Lemma: Suppose $p,q \in \mathbb{C}[t]$ are coprime polynomials and suppose there are four non-zero vectors $(a_1, b_1), \dots, (a_4, b_4) \in \mathbb{C}^2$ defining distinct elements of $\mathbb{P}^1$ such that the polynomials $a_i p + b_i q$ are all squares in $\mathbb{C}[t]$. Then $p$ and $q$ are constant.
Note that since $\lambda \neq 0,1$ the Lemma may be applied here.
