Recall the point-slope formula:
#y-y_1=m(x-x_1)#
Where #m=dy/dx#
When #x=1#, we can plug that into the original equation for #y#
#y=x^3-x^2e^(2x)#
#y=(1)^3-(1)^2e^(2*1)#
#y=1-e^2~~-6.3891#
Finding the derivative of the equation for #y#, we get
#dy/dx=d/dx(x^3)-d/dx(-x^2e^(2x))#
#dy/dx=3x^2-(-x^2 d/dx(e^(2x))+d/dx(-x^2)e^(2x))#
#dy/dx=3x^2+x^2(2e^(2x))+2xe^(2x)#
When #x=1#, the slope of the tangent line is
#dy/dx=3(1)^2+(1)^2(2e^(2(1)))+2(1)e^(2(1))#
#dy/dx=m=3+2e^(2)+2e^(2)=3+4e^2~~32.5562#
Plugging slope, #m=32.5562#, the given #x=1#, and the derived #y=-6.3891#, all into the point slope formula, we get
#y-y_1=m(x-x_1)#
#y+6.3891=32.5562(x-1)#
#y=32.5562x-32.5562-6.3891#
#y=32.5562x-38.9453#
Or, if you need to be exact
#y-1+e^2=(3+4e^2)(x-1)#
#y=3x-3+4xe^2-4e^2+1-e^2#
#y=(3+4e^2)x-(2+5e^2)#
