Let $\Phi$ denote the cumulative distribution function of the standard normal distribution. How can we show that $$\Gamma(a,b):=\Phi\left(-\frac b{2\sqrt a}\right)+e^{\frac{a-b}2}\Phi\left(\frac b{2\sqrt a}-\sqrt a\right)\;\;\;\text{for }(a,b)\in(0,\infty)\times\mathbb R.$$
How can we show that $\Gamma$, $\frac{\partial\Gamma}{\partial b}$ and $\frac{\partial^2\Gamma}{\partial b^2}$ are bounded (in both arguments)?
EDIT: Note that $$\frac{\partial\Gamma}{\partial b}(a,b)=-\frac12e^{\frac{a-b}2}\Phi\left(\frac b{2\sqrt a}-\sqrt a\right)<0\tag1$$ and $$\frac{\partial^2\Gamma}{\partial b^2}(a,b)=\frac14\left(e^{\frac{a-b}2}\Phi\left(\frac b{2\sqrt a}-\sqrt a\right)-\frac1{\sqrt{2\pi a}}e^{-\frac{b^2}{8a}}\right)\tag2$$ for all $(a,b)\in(0,\infty)\times\mathbb R$. By $(1)$, $$\mathbb R\ni b\mapsto\Gamma(a,b)\tag3$$ is decreasing for all $a>0$ and since $$\Gamma(a,b)\le\Phi\left(-\frac b{2\sqrt a}\right)+\frac 2{\sqrt{2\pi a}}e^{-\frac{b^2}{8a}}\xrightarrow{b\to-\infty}1\;\;\;\text{for all }a>0\text{ and }b\le a,\tag4$$ we conclude $$\Gamma(a,b)\le1\;\;\;\text{for all }(a,b)\in(0,\infty)\times\mathbb R.\tag5$$
What can we do for $\frac{\partial\Gamma}{\partial b}$ and $\frac{\partial^2\Gamma}{\partial b^2}$?
Let $f(x)=e^{-x}$ and $g(x)=\frac{e^x}{1+e^x}$, then obviously $f$ and $g$ are both unbounded, so their product is also unbounded. The only problem with this reasoning is that the product is $f(x)\cdot g(x) = \frac{1}{1+e^x}$, which is bounded. Do you see what went wrong here?
Let's go back to your original question. Let fix $b=0$, then the function $e^{a/2}$ is unbounded, however for big $a$, $\Phi(\sqrt{a})$ behaves like $e^{-a/2}$, which compensates for the growth.
In generall if $a-b>0$, then $\frac{b}{2\sqrt a}-\sqrt a\le0$. This means that $\Phi(\frac{b}{2\sqrt a}-\sqrt a)$ can be bounded by something like $e^{\frac{b-a}{2}-\frac{b^2}{8a}}$, which means that the product is bounded. Of course what I've written here is not a proper proof, but this is the essense of the proof. You need to find a precise bound for $\Phi(x)$ when $x<0$ and then check that it decays faster than $e^{\frac{b-a}{2}}$.
Let's assume that $a-b>0$, $\frac{b}{2\sqrt a}-\sqrt a>0$ and $\sqrt a >0$ (this last one is a reasonable assumption, but still it is an assumption that is needed). Then we have $$ \frac{b}{2\sqrt a}-\sqrt a>0 \implies b-2a>0 .$$ Now we take $b-2a>0$ and $a-b>0$, we add them together and we get $-a>0$ which is a contradiction. So $a-b>0$ implies $\frac{b}{2\sqrt a}-\sqrt a \le 0$ (or $\sqrt a <0$ which is a case that needs to be discussed).
This mean that when the "problematic" exponent $a-b$ is positive, $\Phi$ compensates for this growth.
The general method for the derivatives is to establish that they are continuously differentiable and then control the growth to infinity. Because of the exponential convergence to some value, the derivative converges exponentially to 0. This implies that it is bounded.
In your case the issue is simpler. Just differentiate, $\Phi$ has a nice derivative. However, unless there is some relation you haven't mentioned, the derivatives you've written are wrong (I may be wrong here).
You have to be careful what is actually you are asking for. When you differentiate with respect to $b$ you get $\sqrt a$ in the denominator. This means that the derivative is unbounded close to $a=0$. The derivatives are bounded only if you fix $\varepsilon>0$ and ask that $a>\varepsilon$. (this is not true if the derivattives are correct)
I am assuming here that (1) and (2) are correct. In this case they are bounded.
If you know how to prove that $\Gamma$ is bounded, then (1) is also bounded as it is just the problematic term that was discussed above.
For (2) you need to show only that $\frac1{\sqrt{2\pi a}}e^{-\frac{b^2}{8a}}$ is bounded. The only region this can be problematic is $a\to0$ as then $\frac1{\sqrt{2\pi a}}$ goes to infinity "polynomially" fast (with exponent $1/2$). However, if$a\to0$, then $-\frac{b^2}{8a}$ goes to infinity polynomially fast (with exponent 1) so $e^{-\frac{b^2}{8a}}$ goes to $0$ exponentially fast. This dominates and the product goes to $0$.
This last argument follows from the fact that $x^{-1}e^x$ goes to infinity as $x\to\infty$ (you can show this using the Taylor series).
OK, fine, I will write the steps clearly.
First you should check that for all $x<0$ there exists $M>0$ such that $\Phi(x) \le M e^{-\frac{x^2}{2}}$.
Let's define $f(a,b) := e^{\frac{a-b}2}\Phi\left(\frac b{2\sqrt a}-\sqrt a\right)$.
Let's check that $f$ is bounded. We have two cases:
$a-b\le0$ means that $e^{\frac{a-b}2}$ is bounded so $f$ is bounded.
$a-b>0$ means that $\frac{b}{2\sqrt a}-\sqrt a \le 0$ (assuming that $\sqrt a>0$) so we use the bound and get $$ \Phi\left(\frac b{2\sqrt a}-\sqrt a\right) \le M e^{-\frac{b^2}{8a}+\frac{b-a}{2}}. $$ This means that $$ e^{\frac{a-b}2}\Phi\left(\frac b{2\sqrt a}-\sqrt a\right) \le M e^{-\frac{b^2}{8a}}. $$
From the above we deduce that $f$ is bounded on $(0,\infty)\times\mathbb R$.
Observe that $\frac{\partial\Gamma}{\partial b}(a,b)$ is just a constant times $f(a,b)$, which means that $\frac{\partial\Gamma}{\partial b}(a,b)$ is bounded.
Similarly $\frac{\partial^2\Gamma}{\partial b^2}(a,b)$ is a constant times $f(a,b)$ plus a constant times $a^{-1/2}e^{-\frac{b^2}{8a}}$. Let $c=1/a$, then $$a^{-1/2}e^{-\frac{b^2}{8a}} = c^{1/2}e^{-\frac{b^2c}{8}}.$$ Convince yourself that the LHS is bounded for every $a\in(0,\infty)$ if and only if the RHS is bounded for every $c\in(0,\infty)$. For every $b$ the RHS is analytic on $(0,\infty)$, so it is bounded on $(0,N)$ for every $N>0$. Then check that the limit at infinity is 0. This proves that the RHS is bounded.