\(f’\left( x \right) = {\left( {3{e^{{x^2}}}} \right)^\prime } = 3{e^{{x^2}}} \cdot {\left( {{x^2}} \right)^\prime } = 6x{e^{{x^2}}}.\)
\(f’\left( 0 \right) = 6 \cdot 0 \cdot {e^0} = 0;\) \(f\left( 0 \right) = 3 \cdot {e^0} = 3.\)
Тогда:
\(f’\left( x \right)-2xf\left( x \right) + \dfrac{1}{3}f\left( 0 \right)-f’\left( 0 \right) = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,6x{e^{{x^2}}}-2x \cdot 3{e^{{x^2}}} + \dfrac{1}{3} \cdot 3-0 = 1\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,6x{e^{{x^2}}}-6x{e^{{x^2}}} + 1 = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,1 = 1.\)
Что и требовалось доказать.