1. **State the problem:** Given the implicit function defined by the equation $$z(z^2+3x)+3y=0,$$ we need to prove that $$\frac{d^2z}{dx^2}+\frac{d^2z}{dy^2}=\frac{2z(x-1)}{(z^2+x)^3}.$$\n\n2. **Rewrite the equation:** The given equation is \n$$z(z^2+3x)+3y=0 \implies z^3 + 3xz + 3y = 0.$$\n\n3. **Find the first derivatives:** We treat $z$ as a function of $x$ and $y$, i.e., $z=z(x,y)$. Differentiate implicitly with respect to $x$:\n$$\frac{d}{dx}(z^3) + \frac{d}{dx}(3xz) + \frac{d}{dx}(3y) = 0.$$\nUsing the chain and product rules:\n$$3z^2 \frac{\partial z}{\partial x} + 3z + 3x \frac{\partial z}{\partial x} + 0 = 0,$$\nsince $y$ is independent of $x$.\nGroup terms:\n$$\left(3z^2 + 3x\right) \frac{\partial z}{\partial x} = -3z.$$\nTherefore,\n$$\frac{\partial z}{\partial x} = \frac{-3z}{3(z^2 + x)} = \frac{-z}{z^2 + x}.$$\n\n4. **Differentiate implicitly with respect to $y$:**\n$$\frac{d}{dy}(z^3) + \frac{d}{dy}(3xz) + \frac{d}{dy}(3y) = 0.$$\nSince $x$ is independent of $y$,\n$$3z^2 \frac{\partial z}{\partial y} + 3x \frac{\partial z}{\partial y} + 3 = 0,$$\nwhich simplifies to\n$$\left(3z^2 + 3x\right) \frac{\partial z}{\partial y} = -3,$$\nso\n$$\frac{\partial z}{\partial y} = \frac{-3}{3(z^2 + x)} = \frac{-1}{z^2 + x}.$$\n\n5. **Find the second derivatives:**\nDifferentiate $\frac{\partial z}{\partial x} = \frac{-z}{z^2 + x}$ with respect to $x$:\nUse quotient rule or implicit differentiation:\n$$\frac{\partial^2 z}{\partial x^2} = \frac{-(z^2 + x) \frac{\partial z}{\partial x} + z(2z \frac{\partial z}{\partial x} + 1)}{(z^2 + x)^2}.$$\nSubstitute $\frac{\partial z}{\partial x} = \frac{-z}{z^2 + x}$:\nNumerator:\n$$-(z^2 + x) \left(\frac{-z}{z^2 + x}\right) + z \left(2z \left(\frac{-z}{z^2 + x}\right) + 1\right) = z + z \left(-\frac{2z^2}{z^2 + x} + 1\right).$$\nSimplify:\n$$z + z \left(1 - \frac{2z^2}{z^2 + x}\right) = z + z - \frac{2z^3}{z^2 + x} = 2z - \frac{2z^3}{z^2 + x} = \frac{2z(z^2 + x) - 2z^3}{z^2 + x} = \frac{2z x}{z^2 + x}.$$\nTherefore,\n$$\frac{\partial^2 z}{\partial x^2} = \frac{2z x}{(z^2 + x)^3}.$$\n\n6. **Differentiate $\frac{\partial z}{\partial y} = \frac{-1}{z^2 + x}$ with respect to $y$:**\n$$\frac{\partial^2 z}{\partial y^2} = - \frac{\partial}{\partial y} \left(\frac{1}{z^2 + x}\right) = - \left(-\frac{2z \frac{\partial z}{\partial y}}{(z^2 + x)^2}\right) = \frac{2z \frac{\partial z}{\partial y}}{(z^2 + x)^2}.$$\nSubstitute $\frac{\partial z}{\partial y} = \frac{-1}{z^2 + x}$:\n$$\frac{\partial^2 z}{\partial y^2} = \frac{2z \left(-\frac{1}{z^2 + x}\right)}{(z^2 + x)^2} = -\frac{2z}{(z^2 + x)^3}.$$\n\n7. **Sum the second derivatives:**\n$$\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \frac{2z x}{(z^2 + x)^3} - \frac{2z}{(z^2 + x)^3} = \frac{2z(x - 1)}{(z^2 + x)^3}.$$\n\n**Final answer:**\n$$\boxed{\frac{d^2 z}{dx^2} + \frac{d^2 z}{dy^2} = \frac{2z(x - 1)}{(z^2 + x)^3}}.$$