In general, for any non-constant morphism $f:C \to \mathbb P^n$, from a $1$-dimensional Cohen-Macaulay curve $C$, one has that $$H^1(C,f^*T_{\mathbb P^n}\otimes \omega_C)=0.$$

Indeed, (as already pointed out by Al e) considering the pull-back of the Euler sequence $$0 \to f^*\mathscr O_{\mathbb P^n} \to f^*\mathscr O_{\mathbb P^n}(1)^{\oplus (n+1)} \to f^*T_{\mathbb{P}^n} \to 0$$ and using the fact that $H^2(C,\mathscr O_C)=0$ automatically by dimension considerations, it is enough to prove that $$H^1(C, f^*\mathscr O_{\mathbb P^n}(1)\otimes \omega_C)=0.$$ By Serre duality this is dual to $H^0(C,f^*\mathscr O_{\mathbb P^n}(-1))$ and since $f$ is non-constant, this is an anti-ample line bundle on $C$ and hence has no global sections.

Remark: One needs the Cohen-Macaulay condition for Serre duality and so that $\omega_C$ is sensible.

In general, for any non-constant morphism $f:C \to \mathbb P^n$, from a $1$-dimensional Cohen-Macaulay (for instance reduced) curve $C$, one has that $$H^1(C,f^*T_{\mathbb P^n}\otimes \omega_C)=0.$$

Indeed, (as already pointed out by Al e) considering the pull-back of the Euler sequence $$0 \to f^*\mathscr O_{\mathbb P^n} \to f^*\mathscr O_{\mathbb P^n}(1)^{\oplus (n+1)} \to f^*T_{\mathbb{P}^n} \to 0$$ and using the fact that $H^2(C,\mathscr O_C)=0$ automatically by dimension considerations, it is enough to prove that $$H^1(C, f^*\mathscr O_{\mathbb P^n}(1)\otimes \omega_C)=0.$$ By Serre duality this is dual to $H^0(C,f^*\mathscr O_{\mathbb P^n}(-1))$ and since $f$ is non-constant, this is an anti-ample line bundle on $C$ and hence has no global sections.

Remark: One needs the Cohen-Macaulay condition for Serre duality and so that $\omega_C$ is sensible.

In general, for any non-constant morphism $f:C \to \mathbb P^n$, from a $1$-dimensional Cohen-Macaulay curve $C$, one has that $$H^1(C,f^*T_{\mathbb P^n}\otimes \omega_C)=0.$$

Indeed, (as already pointed out by Al e) considering the pull-back of the Euler sequence $$0 \to f^*\mathscr O_{\mathbb P^n} \to f^*\mathscr O_{\mathbb P^n}(1)^{\oplus (n+1)} \to f^*T_{\mathbb{P}^n} \to 0$$ and using the fact that $H^2(C,\mathscr O_C)=0$ automatically by dimension considerations, it is enough to prove that $$H^1(C, f^*\mathscr O_{\mathbb P^n}(1)\otimes \omega_C)=0.$$ By Serre duality this is dual to $H^0(C,f^*\mathscr O_{\mathbb P^n}(-1))$ and since $f$ is non-constant, this is an anti-ample line bundle on $C$ and hence has no global sections.

Remark: One needs the Cohen-Macaulay condition for the Serre duality and so that $\omega_C$ is sensible.

In general, for any non-constant morphism $f:C \to \mathbb P^n$, from a $1$-dimensional Cohen-Macaulay curve $C$, one has that $$H^1(C,f^*T_{\mathbb P^n}\otimes \omega_C)=0.$$

Indeed, (as already pointed out by Al e) considering the pull-back of the Euler sequence $$0 \to f^*\mathscr O_{\mathbb P^n} \to f^*\mathscr O_{\mathbb P^n}(1)^{\oplus (n+1)} \to f^*T_{\mathbb{P}^n} \to 0$$ and using the fact that $H^2(C,\mathscr O_C)=0$ automatically by dimension considerations, it is enough to prove that $$H^1(C, f^*\mathscr O_{\mathbb P^n}(1)\otimes \omega_C)=0.$$ By Serre duality this is dual to $H^0(C,f^*\mathscr O_{\mathbb P^n}(-1))$ and since $f$ is non-constant, this is an anti-ample line bundle on $C$ and hence has no global sections.

Remark: One needs the Cohen-Macaulay condition for Serre duality and so that $\omega_C$ is sensible.