dy/dx=y'=(2/3)x^(2/3-1)=(2/3)x^(-1/3) or 2/3x^(1/3).
The slope of y=(x/3)-5 is 1/3 and this is the slope of all lines parallel to it.
The general equation of such lines is: y=(x/3)+a where a is a constant.
So at a particular point on the curve 1/3=2/(3x^(1/3)), 1=2/x^(1/3), x^(1/3)=2 and x=2^3=8, making y=8^(2/3)=4.
The tangent line is given by y=x/3+a, where a is found by substituting for x and y:
4=8/3+a, so a=4-8/3=4/3 and y=(1/3)(x+4) is the tangent line.