Find where y'=0 given that y=10^xlogx-10x
I presume you know that the derivative of -10x is -10.
10^x log x is a product so use the product rule: (10^x log x)'= (10^x)' log x+ 10^x (log x)'.
The derivative of 10^x is (10^x)'= (10^x) ln x and the derivative of log x is
1) 1/(x ln(10)) if it is common logarithm.
2) 1/x, if it is natural logarithm.