I would like to show that when $f(x,y)$ is irreducible over $\mathbb{C}[x,y]$ then $f(x^2,y)$ is irreducible over $\mathbb{C}[x,y]$. I know that this is not true in general, for example, $f(x,y) = y^2 - x$. However, I know that the original polynomial contains only even powers of $x$. Does this extra condition allow me to conclude $f(x^2,y)$ is irreducible over $\mathbb{C}[x,y]$? 



Irreducibility after substitution
The answer is yes. (In what follows, I write $f(x^2,y)$ for your $f(x,y)$.)


1) By symmetry, if $h(x,y)$ is an irreducible factor of $g(x^2,y)$ then so is $h(-x,y)$. 


2) In particular, if $h(x,y)$ is an irreducible factor of $f(x^4,y)$ then so is $h(-x,y)$.


3) Write $f(x^4,y)$ as a product of irreducible factors. Let $h(x,y)$ be such a factor. 


3a) Suppose that $h(x,y)=\alpha h(-x,y)$ for some constant $\alpha$. Comparing coefficients, we see that either $\alpha=1$ and $h$ is an even function of $x$ or $\alpha=-1$ and $h$ is an odd function of $x$.


3aa) Suppose $\alpha=1$ and $h$ is an even function of $x$. Then $h(\sqrt{x},y)$ is a polynomial factor of $f(x^2,y)$, hence --- by the irreducibility assumption ---$h(\sqrt{x},y)=f(x^2,y)$ up to a constant. Thus $h(x,y)=f(x^4,y)$ up to a constant, and the latter is irreducible.


3ab) Suppose $\alpha=-1$ and $h$ is an odd function of $x$. Then $h$ is divisible by $x$, hence by irreducibility we can assume $h(x,y)=x$. But if $x$ divides $f(x^4,y)$ then the latter has no constant term in $x$, so $x$ divides $f(x^2,y)$, a contradiction. [**Note:** My original post overlooked this case. Many thanks to [Emil Jerabek](https://mathoverflow.net/users/12705/emil-je%C5%99%C3%A1bek) for pointing this out and keeping me honest.]


3b) Suppose instead that $h(x,y)$ is not a constant factor of $h(-x,y)$. Write $H(x^2,y)=h(x,y)h(-x,y)$. Then $f(x^4,y)/H(x^2,y)$ is a polynomial in $x^2$ and $y$, so (1) applies and we proceed by induction to write $f(x^4,y)$ as a product of factors $H\_i(x^2,y)$. 


4) It follows that $f(x^2,y)$ is the product of the $H\_i(x,y)$. By assumption $f(x^2,y)$ is irreducible, so there is only one $H\_i$. 


5) Thus $f(x^4,y)=h(x,y)h(-x,y)$ where $h$ is irreducible. 


6) If $h$ contains terms of odd degree in $x$, let $d$ be the smallest of those odd degrees. Then $f(x^4,y)$ contains a term of degree $2d$ in $x$, a contradiction. So all terms of $h$ are of even degree in $x$.


7) Thus $f(x^2,y)=h(\sqrt{x},y)h(-\sqrt{x},y)$ is a polynomial factorization, contradicting the assumption that $f(x^2,y)$ is irreducible.


