In general, the integrating factor method can solve differential equations of the form
("d"y)/("d"x) + "P"(x)*y = "Q"(x).
Observe that,
"d"/("d"x)(e^(\int "P"(x) "d"x)y) = e^(\int "P"(x) "d"x)*("d"y)/("d"x)+P(x)e^(\int "P"(x) "d"x)y
"d"/("d"x)(e^(\int "P"(x) "d"x)y) = e^(\int "P"(x) "d"x)(("d"y)/("d"x) + "P"(x)*y)
Then,
"d"/("d"x)(e^(\int "P"(x) "d"x)y) = e^(\int "P"(x) "d"x)"Q"(x),
e^(\int "P"(x) "d"x)y = \int e^(\int "P"(x) "d"x)"Q"(x) "d"x.
This is the general solution to the differential equation.
We call e^(\int "P"(x) "d"x) the integrating factor.
In your specific case, "P"(x) = 1, "Q"(x) =e^(3x). The integrating factor is therefore
e^(\int 1 "d"x)=e^(x)
Then, substituting,
e^x*y = \int e^(x)*e^(3x) "d"x
e^x*y = \int e^(4x) "d"x
e^x*y = e^(4x)/4 + C
y = (e^(4x)/4+C)/e^x
y=1/4e^(3x) + Ce^(-x)
So, yes, you are correct. A method to check your answer in the future would be to differentiate y and substitute ("d"y)/("d"x) and y into your differential equation.
