(Ⅰ) First, we compare the two expressions:
$$\frac{x^{2}}{x+y} \quad \text{and} \quad \frac{3x-y}{4}$$.
The difference between them is:
$$\frac{x^{2}}{x+y} - \frac{3x-y}{4} = \frac{4x^2 - (3x-y)(x+y)}{4(x+y)} = \frac{4x^2 - 3x^2 - xy + xy + y^2}{4(x+y)} = \frac{x^2 + y^2}{4(x+y)}$$
Since $x^2$ and $y^2$ are both nonnegative because $x, y > 0$, and $x + y > 0$ by the same argument, we have:
$$\frac{x^2 + y^2}{4(x+y)} \geq 0$$
Therefore,
$$\frac{x^{2}}{x+y} \geq \frac{3x-y}{4}$$.

(Ⅱ) By applying the result from (Ⅰ), we have:
$$\frac{x^{3}}{x+y} \geq \frac{3x^{2}-xy}{4}$$
Similarly, we can derive that:
$$\frac{y^{3}}{y+z} \geq \frac{3y^{2}-yz}{4}$$
$$\frac{z^{3}}{z+x} \geq \frac{3z^{2}-zx}{4}$$
Now, consider the expression $x^2 + y^2 + z^2 - (xy + yz + zx)$:
$$x^2 + y^2 + z^2 - (xy + yz + zx) = \frac{1}{2} \left[(x-y)^2 + (y-z)^2 + (z-x)^2\right] \geq 0$$
This implies that
$$x^2 + y^2 + z^2 \geq xy + yz + zx$$
Hence, combining the inequalities, we get:
$$\frac{x^{3}}{x+y} + \frac{y^{3}}{y+z} + \frac{z^{3}}{z+x} \geq \frac{3x^{2}-xy+3y^{2}-yz+3z^{2}-zx}{4}$$
$$= \frac{3(x^{2}+y^{2}+z^{2}) - (xy+yz+zx)}{4} \geq \frac{3(xy+yz+zx) - (xy+yz+zx)}{4} = \frac{2(xy+yz+zx)}{4}$$
Therefore, we have:
$$\frac{x^{3}}{x+y} + \frac{y^{3}}{y+z} + \frac{z^{3}}{z+x} \geq \frac{xy+yz+zx}{2}$$
And we conclude with:
$$\boxed{\frac{x^{3}}{x+y} + \frac{y^{3}}{y+z} + \frac{z^{3}}{z+x} \geq \frac{xy+yz+zx}{2}}$$