Find dy/dx at (3,8) for 5xy^2/3-x^2y=-12
5xy^2/3 - x^2y = -12
5y^(2/3) + 5x*2/3 y^(-1/3) y' - 2xy - x^2y' = 0
(10x/3y^(1/3)-x^2)y' = 2xy - 5y^(2/3)
y' = (2xy - 5y^(2/3))/(10x/3y^(1/3)-x^2) = [-3y(2x∛y-5)]/[x(3x∛y-10)]
Now just plug in (3,8) for (x,y) and you have
[-3*8(2*3*2-5)]/[3(3*3*2-10)] = -7
so, the tangent line there is y-8 = -7(x-3)
see the graphs at
www.wolframalpha.com/input/?i=plot+5xy%5E(2%2F3)+-+x%5E2y+%3D+-12,+y%3D-7(x-3)%2B8+for+0+%3C%3D+x+%3C%3D+5