Answer
$$27\left( {1 + \ln 3} \right)$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {3 + h} \right)}^{3 + h}} - 27}}{h} \cr
& {\text{rewrite}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {3 + h} \right)}^{3 + h}} - 27}}{h} \cr
& {\text{using the definition of the derivative }}f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} \cr
& {\text{comparing }}\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {3 + h} \right)}^{3 + h}} - 27}}{h}{\text{ with the definition of the derivative}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {3 + h} \right)}^{3 + h}} - 27}}{h}:f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {3 + h} \right) - f\left( 3 \right)}}{h} \cr
& {\text{we can note that }}f\left( x \right) = {x^x}{\text{ and }}a = 3 \cr
& {\text{then}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,f\left( x \right) = {x^x} \cr
& \,\,\,\,\,\,\,\,\ln f\left( x \right) = \ln {x^x} \cr
& \,\,\,\,\,\,\,\,\ln f\left( x \right) = x\ln x \cr
& {\text{differentiate}} \cr
& \,\,\,\,\,\,\,\,\frac{{f'\left( x \right)}}{{f\left( x \right)}} = x\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ x \right] \cr
& \,\,\,\,\,\,\,\,\frac{{f'\left( x \right)}}{{f\left( x \right)}} = x\left( {\frac{1}{x}} \right) + \ln x\left( 1 \right) \cr
& \,\,\,\,\,\,\,\,\frac{{f'\left( x \right)}}{{f\left( x \right)}} = 1 + \ln x \cr
& \,\,\,\,\,\,\,\,f'\left( x \right) = f\left( x \right)\left( {1 + \ln x} \right) \cr
& \,\,\,\,\,\,\,\,f'\left( x \right) = {x^x}\left( {1 + \ln x} \right) \cr
& \,\,\,\,\,\,\,\,f'\left( 3 \right) = {3^3}\left( {1 + \ln 3} \right) \cr
& \,\,\,\,\,\,\,\,f'\left( 3 \right) = 27\left( {1 + \ln 3} \right) \cr
& {\text{and}} \cr
& \,\,\,\,\,\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {3 + h} \right)}^{3 + h}} - 27}}{h} = f'\left( 3 \right) = 27\left( {1 + \ln 3} \right) \cr} $$
