Solve the equation

Case 1: $x<0$ $f(x)=2^x+3^x+6^x-x^2$ is increasing, so the equation $f(x)=0$ has the unique solution $x=-1$

Case2: $x \geq 0$ Assume that there is a solution $s \geq 0$ Then

$s^2 = 2^s + 3^s +6^s \geq 3$

so $s \geq \sqrt{3}$ and hence $\lfloor s \rfloor \geq 1$

But then $s \geq \lfloor s \rfloor$ yields $2^s \geq 2^{\lfloor s \rfloor} = (1+1)^{\lfloor s \rfloor} \geq 1+\lfloor s \rfloor \geq s$

So this means

$6^s>4^s=(2^s)^2 \geq s^2$

Thus $2^s + 3^s +6^s > s^2$ a contradiction

Therefore $x=-1$ is the only solution.

If x<0, all terms on LHS must be fractions, while RHS will always be = or >0. At a look at it we can see x= - 1. What ever x is LHS and RHS > 0.
$So~~x^x+3^x+6^x+1=(1+2^x)(1+3^x)=x+1$ For any x>0, we can see this is not possible since $(1+2^x)>x~ and~ (1+3^x)>1$ . So only value of x is -1.