We can cover $X=\mathbb{P}^2_k \setminus [1:0:0]=D_+(y) \cup D_+(z)$ and the Čech complex is given by $$ 0 \rightarrow \mathcal{O}_X(D_+(y)) \oplus \mathcal{O}_X(D_+(z)) \rightarrow \mathcal{O}_X(D_+(yz) ) \rightarrow 0 $$
where the non-trivial map is $d:(f,g) \mapsto f|_{\mathcal{O}_X(D_+(yz) )}-g|_{\mathcal{O}_X(D_+(yz) )}$
Explicitly we have $\mathcal{O}_X(D_+(y))=k[\frac{x}{y},\frac{z}{y}]$ and $\mathcal{O}_X(D_+(z))=k[\frac{x}{z},\frac{y}{z}]$, while $\mathcal{O}_X(D_+(yz) )=k[\frac{x}{z},\frac{y}{z},\frac{z}{y}]$.
Now $\check{H}^1(X,\mathcal{O}_X)=\mathrm{coker}(d)$, but I have some problem computing it. Any hint?
Let's write $X, Y$ for $\frac{x}{z}, \frac{y}{z}$.
You wish to compute the cokernel of the map $$ k[XY^{-1}, Y^{-1}] \oplus k[X, Y] \mapsto k[X, Y, Y^{-1}] $$ given by $(f, g) \mapsto (f - g)$. Then elements of the form $X^mY^n$ are in the image if and only if either $n > 0$ or $n < -m$. Elements of the form $X^mY^n$ where $-m < n < 0$ therefore generate the cokernel. In fact they are linearly independent in the cokernel (check that no linear combination of them can be in the image of the map), so the cohomology is infinite-dimensional.