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Since we are given the quotient, we do not need long division to find the remainder. Instead, remember that if our remainder is $r(z)$, $$3z^3-4z^2-14z+3=(3z+5)\left(z^2-3z+\frac{1}{3}\right)+r(z).$$Multiplying the divisor and the quotient gives us $$(3z+5)\left(z^2-3z+\frac{1}{3}\right)=3z^3+5z^2-9z^2-15z+z+\frac{5}{3}=3z^3-4z^2-14z+\frac{5}{3}$$Subtracting the above result from the dividend gives us the remainder $$r(z)=3z^3-4z^2-14z+3-\left(3z^3-4z^2-14z+\frac{5}{3}\right)=\boxed{\frac{4}{3}}$$We can make the computation easier by realizing that $r(z)$ is a constant. The constants on both sides must be equal, so $$3=5\cdot \frac{1}{3}+r(z).$$Hence, $r(z)=3-\frac{5}{3}=\frac{4}{3}.$