Generalization of Rational Root Theorem [closed] Can the rational root theorem in univariate var

As said in the comments: the answer is NO in general.
For your particular case, you can do the following thing.
We have
${\displaystyle f(x,y)=-2{(y-\frac{x}{4})}^2+\frac{x^2}{2}+4x^2+x+1}$
${\displaystyle =-2{(y-\frac{x}{4})}^2+\frac{9}{2}{x}^2+x+1}$
${\displaystyle =-2{(y-\frac{x}{4})}^2+\frac{9}{2}{(x+\frac{1}{9})}^2+\frac{17}{18}}$.
Soving ${\displaystyle f(x,y)=0\text{where}x,y\in \mathbb{Q}}$ is the same as solving ${\displaystyle -2u^2+\frac{9}{2}{v}^2=-\frac{17}{18}}$, setting ${\displaystyle u=y-\frac{x}{4}\text{and}V=x+\frac{1}{9}}$ (it is easy to derive x,y from u, v. This is equivalent to
${\displaystyle -36u^2+81v^2=-17}$, that is ${\displaystyle (3u-9v)(3u+9v)=17}$, which is easy to solve. For ${\displaystyle t\in {\mathbb{Q}}^{\times }}$, this is equivalent to solve ${\displaystyle 3u-9v=t,3u+9v=\frac{17}{t}}$. You then get u,v as a rational fraction of t, then you get x,y.
[The idea is that you have an affine conic, so you can reduce it to a "diagonal form", and then reduce to the easier case to a rational conic ${\displaystyle a{x}^2+b{y}^2=c}$. This situation is well understood from quadratic form specialist, and you have systematic methods to solve this even if it can be a bit tricky. Here we are in a particularly nice situation.]

Commenters have pointed to examples of why this won’t generalize. I’ll give a more theoretical reason.
The trick of the rational root theorem is that it takes a polynomial of one variable and converts it to a homogenous polynomial of two variables, and then deduce things about the integer solutions of homogenous polynomial.
For example, rational roots of ${\displaystyle 2x^2+5x+3=0}$ correspond to integer solutions to ${\displaystyle 2x^2+5xy+3y^2=0}$, with ${\displaystyle gcd(x,y)=1}$.From there, we see y must be a factor of ${\displaystyle 2x^2}$. But the GCD condition means y must divide 2. Similarly, x must divide 3.
With multiple variables we can still homogenize. We get, in your equation:
${\displaystyle z^2+xz+(xy-2y^2+4x^2)=0}$
But we can only restrict to integers with ${\displaystyle gcd(x,y,z)=1}$, not pairwise relatively '. For example, if in the original equation, the roots are ${\displaystyle (x,y)=(\frac{1}{2},\frac{1}{3})}$, the roots in the homogenous polynomial would be ${\displaystyle (x,y,z)=(3,2,6)}$.
And, in any event, we don’t know a factor of
${\displaystyle xz+(xy-2x^2+4y^2)}$, so we don’t have a factor of ${\displaystyle z^2}$. Indeed, ${\displaystyle gcd(x,xy-2y^2+4x^2)=gcd(x,2y^2)}$, which are relatively ' if ${\displaystyle gcd(x,2y)=1}$.
So, the heart of it is that homogenous integer polynomials of two variables, x,y, have one term of the form ${\displaystyle a{x}^n}$ and all other terms are divisible by y. There is no other common divisor for the other terms with more than one variable.
But even if there is a common divisor, say:
${\displaystyle 3z^2+2zy+xy=0}$.
you do get ${\displaystyle y\mid 3z^2}$, but you don’t know all the solutions can reduce to an example with ${\displaystyle gcd(y,z)=1}$. So you can’t deduce ${\displaystyle y\mid 3}$.
One final note is that homogeneous polynomials of more than 2 variables need not even have “lead coefficients.” For example:
${\displaystyle x_1^2{x}_2+x_2^2{x}_3+x_3^2{x}_1}$
has no terms of the form ${\displaystyle a{x}_i^3}$.
We can get similar examples with two variables, but they factor:
${\displaystyle x{y}^2+y{x}^2=xy(y+x)}$.