And what is the question?

Would it be: Find the dy/dx?

y = 2^(3x^2)

d/dx a^u = ln(a)*a^u *du/dx

So,

dy/dx = ln(2)*[2^(3x^2)]*6x

dy/dx = 6ln(2)*x*2^(3x^2) ---------answer

dy/dx = 6ln(2) *x *8^(x^2)

dy/dx = ln(2^6) *x *8^(x^2)

dy/dx = ln(64) *x *8^(x^2)

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Or,

y = 2^(3x^2)

y = (2^3)^(x^2)

y = 8^(x^2)

dy/dx = ln(8) *8^(x^2) *2x

dy/dx = 2ln(8) *x *8^(x^2)

dy/dx = ln(8^2) *x *8^(x^2)

dy/dx = ln(64) *x *8^(x^2)