Given the function $f(x)=k x^{2}+k x+k+3$, and for any $x \in[-2,3]$, $f(x) \geqslant 0$ always holds, find the range of real number $k$.

To find the range of $k$ for which $f(x) \geqslant 0$ for all $x \in[-2,3]$, we need to analyze the function $f(x)=k x^{2}+k x+k+3$.

1. **Determine the vertex of the parabola:**
   The function $f(x)$ is a quadratic function, and its graph is a parabola. The vertex of the parabola can be found using the formula $x = -\frac{b}{2a}$, where $a = k$ and $b = k$. Thus, the vertex is at:
   \[
   x = -\frac{k}{2k} = -\frac{1}{2}
   \]

2. **Evaluate $f(x)$ at the vertex:**
   Substitute $x = -\frac{1}{2}$ into $f(x)$:
   \[
   f\left(-\frac{1}{2}\right) = k \left(-\frac{1}{2}\right)^2 + k \left(-\frac{1}{2}\right) + k + 3 = k \cdot \frac{1}{4} - k \cdot \frac{1}{2} + k + 3 = \frac{k}{4} - \frac{k}{2} + k + 3 = \frac{k}{4} + \frac{2k}{4} + 3 = \frac{3k}{4} + 3
   \]
   For $f(x) \geqslant 0$ at the vertex, we need:
   \[
   \frac{3k}{4} + 3 \geqslant 0 \implies \frac{3k}{4} \geqslant -3 \implies 3k \geqslant -12 \implies k \geqslant -4
   \]

3. **Evaluate $f(x)$ at the endpoints of the interval $[-2, 3]$:**
   - At $x = -2$:
     \[
     f(-2) = k(-2)^2 + k(-2) + k + 3 = 4k - 2k + k + 3 = 3k + 3
     \]
     For $f(-2) \geqslant 0$, we need:
     \[
     3k + 3 \geqslant 0 \implies 3k \geqslant -3 \implies k \geqslant -1
     \]

   - At $x = 3$:
     \[
     f(3) = k(3)^2 + k(3) + k + 3 = 9k + 3k + k + 3 = 13k + 3
     \]
     For $f(3) \geqslant 0$, we need:
     \[
     13k + 3 \geqslant 0 \implies 13k \geqslant -3 \implies k \geqslant -\frac{3}{13}
     \]

4. **Combine the conditions:**
   The conditions from the vertex and the endpoints are:
   \[
   k \geqslant -4, \quad k \geqslant -1, \quad k \geqslant -\frac{3}{13}
   \]
   The most restrictive condition is $k \geqslant -\frac{3}{13}$.

Therefore, the range of $k$ is:
\[
\boxed{k \geqslant -\frac{3}{13}}
\]
Solution Method 1: (1) When $k=0$, $f(x)=3 \geqslant 0$ always holds in $[-2,3]$;
(2) When $k>0$, $f(x)=k\left(x^{2}+x+1\right)+3 \geqslant 0$ always holds in $[-2,3]$;
(3) When $k<0$, since $-\frac{1}{2} \in[-2,3]$, so $f(x)=k\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4} k+3 \geqslant 0$ always holds in $[-2,3]$ $\Leftrightarrow\left\{\begin{array}{l}f(-2) \geqslant 0 \\ f(3) \geqslant 0\end{array} \Leftrightarrow\left\{\begin{array}{l}k \geqslant-1 \\ k \geqslant-\frac{3}{13} .\end{array}\right.\right.$ Therefore, $-\frac{3}{13} \leqslant k<0$. In summary, $k \geqslant-\frac{3}{13}$.
Method 2: From $f(x) \geqslant 0$ we get $k \geqslant \frac{-3}{x^{2}+x+1}, x \in[-2,3]$, which is equivalent to $k \geqslant$ $\left(\frac{-3}{x^{2}+x+1}\right)_{\max }=-\frac{3}{13}$.