I need to show that $$f=x^2y^5+x^3+xy^2+y^2+x-1$$ is irreducible in $\mathbb{C}[x,y]$.
What I did so far is "watching" $f$ as a polynomial in $\Bbb{C}[x]$ with coefficients in $\Bbb{C}[y]$ (since $A[x][y]\simeq A[x,y]$, where $A$ is a ring), so we can rewrite $f$ as
$$f=a_3(y)x^3+a_2(y)x^2+a_1(y)x+a_0(y)$$ where $$a_3(y)=1,\ a_2(y)=y^5,\ a_1(y)=y^2+1,\ a_0(y)=y^2-1.$$
So, since $\deg(f)=3$, to show $f$ is irreducible, it is sufficient to show that $f$ has no roots in $\Bbb{C}[y]$. Furthermore, if $f$ has any roots, they must divide the independent term $a_0(y)=y^2-1$. Then the possible roots are
$$\ \pm(y+1),\ \pm(y-1),\ \pm(y^2-1).$$
Now I should plug those values on $f$ and check that none of them are roots of $f$, but I'm not sure if this is the right way to conclude that. I did
$$f(y+1)=y^7+2y^6+2y^3+5y^2+4y+1.$$
Can I conclude then that $f(y+1)\neq 0$ and discard $y+1$ as a possible root of $f$? I'm still having some difficulties regarding polynomials in multiple variables, so excuse me if this is a naive question. Any help would be appreciate!
As you noticed $$f=x^3+a_2(y)x^2+a_1(y)x+a_0(y)\in\mathbb C(y)[x],$$ where $$a_2(y)=y^5,\ a_1(y)=y^2+1,\ a_0(y)=y^2-1.$$ Suppose that $f$ is reducible and find $u(y)\in\mathbb C[y]$ such that $f(u(y))=0$. So $$u^3(y)+y^5u^2(y)+(y^2+1)u(y)+y^2-1=0.$$ Then $u(y)\mid y^2-1$, so $\deg u(y)\le2$. But in this case $\deg y^5u^2(y)>\deg u^3(y)$, $\deg y^5u^2(y)>\deg (y^2+1)u(y)$, and $\deg y^5u^2(y)>\deg (y^2-1)$, so the above equality is not possible.