See answers below for a proof of a special case to the following conjecture.
I am looking to prove the following conjecture:
Conjecture: Let $0<b<a<b+1$. Then, $$ f(z):=-\frac{\mathrm d}{\mathrm db}{_2F_1}\left({b-a,b-a\atop b};z\right) $$ is positive and strictly increasing on $z\in(0,1)$.
I have successfully shown for all $a,b>0$ and $z\in(0,1)$ that $\operatorname{sign}(f(z))=\operatorname{sign}(b-a)$, which gives the necessary result for the first part of the conjecture. However, I have been unable to establish monotonicity in $z$ for the full domain of $a$. So is this conjecture indeed true? The proof below shows that $f>0$ for $b<a<b+1$ and I believe it contains clues for establishing that $f$ is also increasing in $z$ (it does...see answer below for special case).
Proof that $f>0$ for $0<b<a<b+1$:
Using DLMF 15.8.1 we write $$ \begin{aligned} f(z) &=-\partial_b F(b-a,b-a;b;z)\\ &=-\partial_b (1-z)^{2a-b}F(a,a;b;z)\\ &=(1-z)^{2a-b}(Q_2-Q_1), \end{aligned} $$ where $Q_1=-\log(1-z)F(a,a;b;z)$ and $Q_2=-\partial_b F(a,a;b;z)$. Upon inspection of their power series expansions in $z$, both $Q_1$ and $Q_2$ have nonnegative coefficients and thus are positive on $z\in(0,1)$. Therefore, all we need to show is is $Q_1<Q_2$. Using the series expansion for the logarithm and hypergeometric function we find for the product $$ Q_1=\sum_{k=1}^\infty\frac{(a)_k^2}{(b)_k}\sum_{\ell=0}^{k-1}\frac{(a+k)_{-\ell-1}^2}{(b+k)_{-\ell-1}(k-\ell-1)!}\frac{1}{\ell+1}z^k, $$ where $(a)_n:=\Gamma(a+n)/\Gamma(a)$ is the Pochhammer symbol (rising factorial). The series coefficients are positive and $(x+k)_{-\ell-1}$ is a positive decreasing function in $x$ on $x\in\Bbb R^+$. Since $0<b<a$ it follows that $0<(a+k)_{-\ell-1}<(b+k)_{-\ell-1}$ and $$ \begin{aligned} Q_1% &<\sum_{k=1}^\infty\frac{(a)_k^2}{(b)_k}\sum_{\ell=0}^{k-1}\frac{(b+k)_{-\ell-1}}{(k-\ell-1)!}\frac{1}{\ell+1}z^k\\ &=\sum_{k=1}^\infty\frac{(a)_k^2}{(b)_k\,k!}(\psi(b+k)-\psi(b))z^k\\ &=Q_2, \end{aligned} $$ with $\psi(z):=\partial_z\log\Gamma(z)$ is the digamma function. The proof is now complete. $\quad\square$
$\newcommand{\pFq}[5]{{_{#1}F_{#2}}\left({#3\atop #4};#5\right)}$ I was able to find a proof that holds for the special case: $b+\color{red}{1/2}\leq a<b+1$. As you will see in the proof, there is only one step that introduces the restriction $b+1/2\leq a$ and my hope is someone can improve on this step to prove the conjecture in the affirmative. I will point out this step in the proof by highlighting it in red. Intuition tells me this may come in handy for a proof of the general case of the conjecture. We need two lemmas and then the proof will follow.
Proof: The proof is conducted by minimizing $g(a)=(x+k+a)_{-\ell-1}(x+k+1-a)_{-\ell-1}$ on $a\in[0,1]$. One can easily show that $g^\prime(1/2)=0$ and $g^{\prime\prime}(1/2)>0$. $\quad\square$
Proof: Using the definition of $f(z)$ and the hypergeometric transformation of DLMF 15.8.1 we have $$ \begin{aligned} \partial_z f(z)% &=-\partial_b\partial_z\pFq{}{}{b-a,b-a}{b}{z}\\ &=-\partial_b\frac{(b-a)^2}{b}\pFq{}{}{b+1-a,b+1-a}{b+1}{z}\\ &=-\partial_b\frac{(b-a)^2}{b}(1-z)^{2a-b-1}\pFq{}{}{a,a}{b+1}{z}\\ &=\cdots\\ &=\frac{a-b}{b}(1-z)^{2a-b-1}\left(\left(1+\frac{a}{b}\right)\pFq{}{}{a,a}{b+1}{z}+(a-b)(Q_2-Q_1)\right), \end{aligned} $$ where $Q_1=-\log(1-z)F(a,a;b+1;z)$ and $Q_2=-\partial_b F(a,a;b+1;z)$. Using the hypergeometric series, logarithmic series, and Cauchy product of power series we have after some algebraic manipulation $$ Q_1=\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_k}\sum_{\ell=0}^{k-1}\frac{(a+k)_{-\ell-1}^2}{(b+1+k)_{-\ell-1}(k-\ell-1)!(\ell+1)}z^k $$ and $$ Q_2=\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_k}\frac{\psi(b+1+k)-\psi(b+1)}{k!}z^k. $$ Both $Q_1$ and $Q_2$ have positive coefficients so that $Q_1,Q_2>0$ on $z\in(0,1)$. Furthermore, noting that $(x+k)_{-\ell-1}$ is positive and decreasing in $x$ one can show that $Q_2-Q_1<0$. Thus, our goal will be to bound $\partial_z f$ from below by bounding $Q_2-Q_1$ from below and then showing that this lower bound for $\partial_z f$ is positive.
$\color{red}{\text{If we restrict $b+1/2\leq a<b+1$}}$ then we may use the monotony of $(x+k)_{-\ell-1}$ and Lemma 1 to write $$ \begin{aligned} Q_1% &<\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_k}\sum_{\ell=0}^{k-1}\frac{(b+k)_{-\ell-1}(b+k+1)_{-\ell-1}}{(b+1+k)_{-\ell-1}(k-\ell-1)!(\ell+1)}z^k\\ &=\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_k k!}(\psi(b+k)-\psi(b))z^k\\ &=:Q_1^\ast. \end{aligned} $$ It follows using the recurrence relation for the digamma function that $$ \begin{aligned} 0 &>Q_2-Q_1\\ &>Q_2-Q_1^\ast\\ &=\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_kk!}\left(\frac{1}{b+k}-\frac{1}{b}\right)z^k\\ &=-\frac{a^2 z}{b(b+1)^2}\sum_{k=0}^\infty\frac{(a+1)_k^2}{(b+2)_kk!}\frac{b+1}{b+k+1}z^k\\ &=-\frac{a^2 z}{b(b+1)^2}\pFq{3}{2}{b+1,a+1,a+1}{b+2,b+2}{z}. \end{aligned} $$
Now using the integral representation in DLMF 16.5.2 we write $$ \begin{aligned} Q_2-Q_1^\ast% &=-\frac{1}{b}\int_0^1 t^b \frac{a^2 z}{b+1}\pFq{}{}{a+1,a+1}{b+2}{zt}\,\mathrm dt\\ &=-\frac{1}{b}\int_0^1 t^b \partial_t\pFq{}{}{a,a}{b+1}{zt}\,\mathrm dt\\ &=\text{IBP}\cdots\\ &=-\frac{1}{b}\pFq{}{}{a,a}{b+1}{z}+\int_0^1 t^{b-1} \pFq{}{}{a,a}{b+1}{zt}\,\mathrm dt\\ &=-\frac{1}{b}\pFq{}{}{a,a}{b+1}{z}+\frac{1}{b}\pFq{3}{2}{b,a,a}{b+1,b+1}{z}, \end{aligned} $$ Substituting this result into the expression for $\partial_z f$ and simplifying we find $$ \begin{aligned} \partial_z f% &>\frac{a-b}{b}(1-z)^{2a-b-1}\left(\left(1+\frac{a}{b}\right)\pFq{}{}{a,a}{b+1}{z}+(a-b)(Q_2-Q_1^\ast)\right)\\ &=\frac{a-b}{b}(1-z)^{2a-b-1}\left(2\pFq{}{}{a,a}{b+1}{z}+\frac{a-b}{b}\pFq{3}{2}{b,a,a}{b+1,b+1}{z}\right)\\ &>0, \end{aligned} $$ since all terms are positive. $\quad\square$
All that is needed is to correct the one step that introduces the restriction on $a$.