Clarification on the solution set of two surfaces being a hyperbola

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$xy=e$ is a hyperbola. Looking at the LHS and RHS we have a hyperbolic parabaloid (a conoid) and a constant. Taken together, the equation yields a hyperbola. Manipulating algebraically, gives $xy=e^{\big(\frac{1}{\ln(xy)}\big)}.$ So we have a hyperbolic paraboloid set equal to some function, yielding a hyperbola.

The intersection between the two unknown functions $\phi(xy)=\phi(1-xy)=e^{\big(\frac{1}{\ln(xy)}\big)}=e^{\big(\frac{1}{\ln(1-xy)}\big)}$ is a hyperbola, namely $2xy=1.$

Q: Does this imply that $x,y$ is algebraic along this intersection but the value of $\phi(xy)$ is unknown?

For example in a lower dimensional case:

$\phi(x)=e^{\big(\frac{1}{\ln(x)}\big)}=\phi(1-x)\implies x=1/2.$ Which is obviously an algebraic number. However, $\phi(1/2)$ is unknown. So it seems that the independent variables are algebraic in these cases but the output is an unknown number.