From the wording of your post, you seem to ask whether there are functions $f_0$ and $f_1$, which have only two slots f0[slot1,slot2]
, such that when composed with the functions $l_0$ and $l_1$, which are functions with 4 variables {λ0,λ1,μ0,μ1}
, return the given values.
Let us use Mathematica to write the rhs of $f_0$ and $f_1$ in terms of $l_0$, $l_1$, $μ_0$, and $μ_1$. That is, let us move from the space where {λ0,λ1,μ0,μ1}
are the variables to the space where {l0,l1,μ0,μ1}
are the variables.
fromλToL = Solve[{l0 == Sqrt[(λ1*(λ0 + μ0))/(λ0*(λ1 + μ1 - 1))],
l1 == Sqrt[(λ0*(λ1 + μ1))/(λ1*(λ0 + μ0 - 1))]}, {λ0, λ1}][[1]];
FullSimplify[{f0[l0, l1] == 1/Sqrt[(λ0 + μ0)/λ0],
f1[l0, l1] == 1/Sqrt[(λ0 + λ0 - 1)/λ0]} /.
fromλToL, Assumptions -> l0 > 0 && l1 > 0]

It becomes obvious now that $f_0$ and $f_1$ cannot be functions of just $l_0$ and $l_1$.
Another reasoning that leads to the same answer: If there were those two functions $f_0$ and $f_1$ then the composition with the two functions $l_0$ and $l_1$ yield functions of {λ0,λ1,μ0,μ1}
. Take derivatives of $f_0(l_0(λ0,λ1,μ0,μ1),l_1(λ0,λ1,μ0,μ1))=rhs0$ and $f_1(l_1(λ0,λ1,μ0,μ1))=rhs1$ respect to λ0,λ1,μ0, and μ1. You will get 8 differential equations. You can then prove (by solving for the derivatives and then replacing those derivatives in the remaining equations) that such a system of equations has no solution.
sol1 = Solve[{l0 == 1/Sqrt[(\[Lambda]0 + \[Mu]0)/\[Lambda]0], l1 == 1/Sqrt[(\[Lambda]0 + \[Mu]0 - 1)/\[Lambda]0]}, {\[Lambda]1, \[Mu]1}]
andSimplify[{l0, l1} /. sol1]
for a start. $\endgroup$