可以用C语言展示下实现其功能的代码吗？

C语言本身不支持图形绘制，需要使用外部库来实现绘制CIE XYZ色域图的功能。这里提供一个使用Graphics.h库绘制CIE XYZ色域图的示例代码，需要使用Turbo C或Borland C等支持Graphics.h库的编译器来编译运行。

#include <graphics.h>

int main()
{
    int gd = DETECT, gm;
    initgraph(&gd, &gm, "");

    // 绘制三角形顶点
    int x1 = 100, y1 = 100;
    int x2 = 300, y2 = 100;
    int x3 = 200, y3 = 300;
    setfillstyle(SOLID_FILL, RED);
    fillellipse(x1, y1, 5, 5);
    setfillstyle(SOLID_FILL, GREEN);
    fillellipse(x2, y2, 5, 5);
    setfillstyle(SOLID_FILL, BLUE);
    fillellipse(x3, y3, 5, 5);

    // 绘制等色线
    setcolor(WHITE);
    for (int i = 0; i <= 100; i += 10) {
        int x4 = x1 + (x2 - x1) * i / 100;
        int y4 = y1 + (y2 - y1) * i / 100;
        int x5 = x2 + (x3 - x2) * i / 100;
        int y5 = y2 + (y3 - y2) * i / 100;
        int x6 = x3 + (x1 - x3) * i / 100;
        int y6 = y3 + (y1 - y3) * i / 100;
        moveto(x4, y4);
        lineto(x5, y5);
        lineto(x6, y6);
        lineto(x4, y4);
    }

    // 绘制等亮度线
    for (int i = 0; i <= 100; i += 10) {
        int x4 = x1 + (x2 - x1) * i / 100;
        int y4 = y1 + (y2 - y1) * i / 100;
        int x5 = x2 + (x3 - x2) * i / 100;
        int y5 = y2 + (y3 - y2) * i / 100;
        int x6 = x3 + (x1 - x3) * i / 100;
        int y6 = y3 + (y1 - y3) * i / 100;
        int x7 = x1 + (x2 - x1) * i / 100 + (x3 - x1) * i / 100;
        int y7 = y1 + (y2 - y1) * i / 100 + (y3 - y1) * i / 100;
        moveto(x4, y4);
        lineto(x7, y7);
        lineto(x5, y5);
        lineto(x6, y6);
        lineto(x4, y4);
    }

    getch();
    closegraph();
    return 0;
}

注意，这只是一个简单的示例代码，实际的绘图需要根据具体的数据和算法进行实现。