Prove that given projective curve has genus $1$. I was working on the exercise 13.21 from Gathmann's notes.

Show that
$$X=\{((x_0:x_1),(y_0:y_1)):(x_0^2+x_1^2)(y_0^2+y_1^2)=x_0x_1y_0y_1)\}\subseteq \mathbb{P}^1\times\mathbb{P^1}$$
is a smooth curve of genus 1.

I can prove it with the following reasoning.
Using the Segre Embedding, the curve consists on all the elements $(x:y:z:w)\in\mathbb{P}^3$ satisfying the equations
$$x^2+y^2+z^2+w^2-xw=0,xw-yz=0,$$
i.e. the vanishing set of $I=(x^2+y^2+z^2+w^2-xw,xw-yz)$.
I can use the Jacobian criterion to prove it's smooth; no issue.
To find the genus, since I'm better at it, I decided to compute the arithmetic genus. To do that, I proved that
$$\{x^2+y^2+z^2+w^2-xw,xw-yz\}$$
is in fact a Gröbner basis (Using GRevLex ordering), so
$$LT(I)=(x^2,yz).$$
Then every minimal free resolution of the quotient over a monomial ideal (I think I can remove the hypothesis of it being monomial; I'm not completely sure, but in such case I don't need to compute $LT(I)$ or even prove that I have a Gröbner basis) generated by two elements has the form $0\to S\to S^2\to S\to 0$, in particular, in this case it has the form
$$0\to S(-4)\to S(-2)^2\to S\to 0$$
which allows me to compute the Hilbert polynomial of $X$, by the method in the first section of The Geometry of Syzygies from Eisenbud (I remember it's also used in Cox's), and with the Hilbert polynomial I also have that the arithmetic genus is $1$.
But can I prove it without computing the Hilbert polynomial?
 A: The quickest way to do this is probably the adjunction formula. The canonical bundle of a curve of bidegree $(a,b)$ is $\mathcal{O}_C(a-2,b-2)$, which simultaneously has degree $2g-2$ and $b(a-2)+a(b-2)$. So $g=(a-1)(b-1)$ in general, and in your case, $g=1$.
Another way to see this (though I'm hesitant to call it a proof) is to visualize what the solutions to this equation look like. Looking in the $\Bbb A^2\subset \Bbb P^1\times\Bbb P^1$ given by $D(x_0)\times D(x_1)$, our curve is cut out by $(1+x^2)(1+y^2)=xy$. Making the substitution $x=ix,y=iy$ this transforms to $(1-x^2)(1-y^2)=-xy$, which has the following graph (courtesy Desmos): 
It's not so difficult to see that when you connect these up inside $\Bbb P^1\times\Bbb P^1$ you get two concentric circles, which as you vary the imaginary parts of $x,y$ behave just like what happens when you slice a torus with a plane. I mention this because of Gathmann's digression earlier in the section about topological Euler characteristic and genus.
A: Based on your response to @KReiser, I'm going to go ahead and show you how to do this with adjunction. The adjunction formula computes the canonical class of a hypersurface from the canonical class of the ambient variety; specifically, it says that for $D\subset X$ a divisor, we have $K_D = (K_X + D)|_D$.
Now $\mathbb P^1  \times \mathbb P^1$ has Picard rank $2$, so we typically specify a divisor class by an ordered pair $(a,b)$. One way of thinking about these classes (at least in the effective case) is that a divisor of type $(a,0)$ is the pullback of a degree $a$ divisor (e.g. the sum of $a$ points) from the first $\mathbb P^1$ factor, and similarly for type $(0,b)$. These pullbacks are simply disjoint unions of lines (all in the same ruling), the fibers over the divisors on $\mathbb P^1$. Taking the union of a divisor of each type, you get a divisor of type $(a,b)$, where now the $a$ lines in one ruling intersect the $b$ lines from the other ruling.
The other thing to know is the intersection pairing: $(1,0)\cdot (0,1) = 1$ (two lines, one from each ruling, intersect in a unique point), and $(1,0)\cdot (1,0) = (0,1)\cdot (0,1) = 0$  (two lines, both from the same ruling, never intersect). Everything else extends by linearity, so in general $(a,b)\cdot (c,d) = ad + bc$.
To use adjunction here, we need $K_{\mathbb P^1 \times \mathbb P^1}$ and $D$. Now hopefully you know that $K_{\mathbb P^1} = -2$; it follows from general facts about products of varieties that $K_{\mathbb P^1 \times \mathbb P^1} = (-2,-2)$. Since your curve is given by an equation of bidegree $(2,2)$, its divisor class is also $(2,2)$. So $K_{\mathbb P^1 \times \mathbb P^1} + D = (-2,-2) + (2,2) = (0,0)$. To compute the degree of the restriction of this class to your curve $D$, we would normally now compute $(0,0)\cdot (2,2) = 0$, but this is unnecessary since the trivial class always restricts to the trivial class. Now since the only curves with trivial canonical class are genus $1$, we are done.
