find the largest number

$$3.14^{3.14}\lt 3.2^{3.5}=3.2^3\times 3.2^{1/2}\lt 3.2^3\times 4^{1/2}=32.768\times 2\lt 81= 3^4=(1/3)^{-4}$$

mathlove's solution is good. I think the following is a bit easier:

$3.14^{3.14} = 3^3\times\left(\frac{3.14}{3}\right)^3\times3.14^{0.14}$. So we just need the product of the second and third factors to be no more than 3.

The second factor is $<1.1^3=1.331<1.5$. The third factor is (much) less than $4^{0.5}=2$. So the product of these factors is $<3$.