What type of function satisfies $f(x,y)=f\left(-x,\frac{1}{y}\right)$? Is there a name for functions satisfying the condition below?

$f(x,y)=f\left(-x,\frac{1}{y}\right)$

 A: Notice that both functions
$$x\mapsto \iota_1(x):=-x,\hspace{.5cm} y\mapsto \iota_2(y):=\tfrac{1}{y},$$
are involutions: 
$$\iota_1(\iota_1(x))=-(-x)=x,\hspace{.5cm} \iota_2(\iota_2(y))=\tfrac{1}{\left(\tfrac{1}{y}\right)}=y.$$
Now whenever you have an operation $\iota$ with $\iota(\iota(x))=x$, which for $\mathbb R\to \mathbb R$ are in fact all functions which mirrored with respect to the $45°$ axis, then for any function $f$, the function $$\hat{f}(x):=f(x)+f(\iota(x)),$$ with $+$ commutative, fulfills $$\hat{f}(\iota(x))=f(\iota(x))+f(\iota(\iota(x)))=f(\iota(x))+f(x)=f(x)+f(\iota(x))=\hat{f}(x)$$
Actually, via the identity
$$f(x)=\tfrac{1}{2}\left(f(x)+f(\iota(x))\right)+\tfrac{1}{2}\left(f(x)-f(\iota(x))\right),$$
all functions have a part which fullfills the relation, except the ones which fulfill the anti-relation $f(\iota(x))=-f(x)$ for which that part is zero. The above works for any number of arguments. 
Take for example $f(x,y):=x\ \text e^y$, which can be split into $$f(x)=\tfrac{1}{2}\left(x\ \text e^y+(-x)\ \text e^\frac{1}{y}\right)+\tfrac{1}{2}\left(x\ \text e^y-(-x)\ \text e^\frac{1}{y}\right).$$
The first part (which we would write as the function $\propto x\left(\text e^y-\text e^\frac{1}{y}\right)$) clearly does the trick, as $x\mapsto -x$ amonts to flipping the two exponentials, which which are also flip.
The function $\cos(x)\left(y^2+\tfrac{1}{y}\right)$ is the first part of $2\cos(x)y^2$. Remark: "It's even with respect to $x$" isn't properly true, as the condition on demands you to do something with $y$ too.
