Multiply top and bottom by 10^x: y=10^x/(10^2x+1); (10^2x)y+y-10^x=0 (multiplying through by 10^2x+1 and rearranging terms); let z=10^x, then yz^2-z+y=0, which is a quadratic in z, solution: z=(1+sqrt(1-4y^2))/2y. Therefore, 10^x=(1+sqrt(1-4y^2))/2y and x=log((1+sqrt(1-4y^2))/2y).
Example 1: y=1/2, x=log(1/1)=0; put x=0 in original equation: y=1/2.
Example 2: Put x=1 in original equation: y=1/(10+0.1)=1/10.1; y^2=1/102.01; 1-4y^2=(99/101)^2; x=log((1+99/101)*10.1)/2)=log((10.1+99/10)/2)=log((10.1+9.9)/2)=log(10) or log(0.1)=1 or -1. So x=log(1+sqrt(1-4y^2))/2y. This shows that there can be two possible values for x for one value of y. This can be seen by the symmetry of the denominator in the original equation. If y=f(x) then f(x)=f(-x), the only unique value being when x=0 and y=1/2.