Identify the general term in the binomial expansion.

The general term $T_{k+1}$ in the expansion of $(a+b)^n$ is given by the formula:

$$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$

In this problem, we have $(Y^2X^2 + X^4)^{45}$. So, $a = Y^2X^2$, $b = X^4$, and $n = 45$.

$$T_{k+1} = \binom{45}{k} (Y^2)^{45-k} (X^2)^{45-k} (X^4)^k$$ $$T_{k+1} = \binom{45}{k} Y^{2(45-k)} X^{2(45-k)} X^{4k}$$ $$T_{k+1} = \binom{45}{k} Y^{90-2k} X^{90-2k+4k}$$ $$T_{k+1} = \binom{45}{k} Y^{90-2k} X^{90+2k}$$

We are looking for terms in $X^{100}$, so we set the exponent of $X$ to 100:

$$90+2k = 100$$ $$2k = 100 - 90$$ $$2k = 10$$ $$k = \frac{10}{2}$$ $$k = 5$$

Since $k=5$ is an integer and $0 \le 5 \le 45$, this is a valid value for $k$.

$$T_{5+1} = T_6 = \binom{45}{5} Y^{90-2(5)} X^{90+2(5)}$$ $$T_6 = \binom{45}{5} Y^{90-10} X^{90+10}$$ $$T_6 = \binom{45}{5} Y^{80} X^{100}$$

$$\binom{45}{5} = \frac{45!}{5!(45-5)!} = \frac{45!}{5!40!} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1}$$ $$\binom{45}{5} = \frac{45}{5 \times 3} \times \frac{44}{4 \times 2} \times 43 \times 42 \times 41$$ $$\binom{45}{5} = (3) \times (11) \times 43 \times (7) \times 41$$ $$\binom{45}{5} = 9 \times 11 \times 43 \times 7 \times 41$$ $$\binom{45}{5} = 99 \times 301 \times 41$$ $$\binom{45}{5} = 29799 \times 41$$ $$\binom{45}{5} = 1,221,759$$

The term containing $X^{100}$ is:

$$1,221,759 Y^{80} X^{100}$$